bases of functional analysischemin/pdf/4m005_2018_w.pdf · 2 normed spaces, banach spaces 27 ......

BASES OF FUNCTIONAL ANALYSIS

Jean-Yves CHEMIN

Laboratoire J.-L. Lions, Campus Pierre et Marie Curie, Case 187

Sorbonne Université, 4 Place Jussieu

75230 Paris Cedex 05, France

Fax: +33 1 44 27 72 00, e-mail: [email protected]

January 2, 2019

Contents

1 Metric spaces 7

1.1 Definition of metric spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.2 Complete spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141.3 Compactness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

2 Normed spaces, Banach spaces 27

2.1 Definitions of normed spaces and Banach spaces . . . . . . . . . . . . . . . . . . 272.2 Spaces of continuous linear maps . . . . . . . . . . . . . . . . . . . . . . . . . . 322.3 Banach spaces, compactness and finite dimension . . . . . . . . . . . . . . . . . 382.4 Compactness in the space of continuous functions: Ascoli’s theorem . . . . . . . 402.5 About the Stone-Weierstrass theorem . . . . . . . . . . . . . . . . . . . . . . . . 412.6 Notions on separable metric spaces . . . . . . . . . . . . . . . . . . . . . . . . . 46

3 Duality in Banach spaces 51

3.1 A presentation of the concept of duality . . . . . . . . . . . . . . . . . . . . . . 513.2 Identifying a normed space as a dual space . . . . . . . . . . . . . . . . . . . . . 543.3 A weaker sense for convergence in E

0 . . . . . . . . . . . . . . . . . . . . . . . . 57

4 Hilbert spaces 63

4.1 Orthogonality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 634.2 Properties of Hilbert spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 654.3 Duality in Hilbert spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 704.4 The adjoint of an operator, self-adjoint operators . . . . . . . . . . . . . . . . . 73

5 Lp

spaces 81

5.1 Measure theory and definition of the Lp spaces . . . . . . . . . . . . . . . . . . 82

5.2 The Lp spaces are Banach spaces . . . . . . . . . . . . . . . . . . . . . . . . . . 84

5.3 Density in Lp spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90

5.4 Convolution and smoothing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 975.5 Duality between L

p and Lp0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103

6 The Dirichlet problem 109

6.1 A classical approach to the problem . . . . . . . . . . . . . . . . . . . . . . . . 1106.2 The concept of quasi-derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . 1116.3 The space H

10 (⌦) and the Dirichlet problem . . . . . . . . . . . . . . . . . . . . 113

3

7 The Fourier transform 117

7.1 The Fourier transform on L1(Rd) . . . . . . . . . . . . . . . . . . . . . . . . . . 117

7.2 The inversion formula and the Fourier-Plancherel theorem . . . . . . . . . . . . 1217.3 Proof of Rellich’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123

8 Tempered distributions in dimension 1 125

8.1 Definition of tempered distributions and examples . . . . . . . . . . . . . . . . 1268.2 Operations on tempered distributions . . . . . . . . . . . . . . . . . . . . . . . . 1318.3 Two applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142

4

Introduction

These are the lecture notes for the “Bases of Functional Analysis” course, part of the 1st yearof the Master’s degree in Mathematics at Pierre & Marie Curie University, Paris. The aimof the course is for the student to acquire an elementary and solid mastery of tools whichare fundamental in mathematics, both as a core subject (geometry, probability theory, partialdifferential equations) and in its applications to physics, mechanics, the study of large-scalesystems, imagery, statistics, etc.

First, let us make some general remarks about these notes. It is not a treatise, in that someclassical results, like the Cauchy-Lipschitz theorem, are absent, having been taught in othercourses, while others, like the Stone-Weierstrass theorem, are stated in a specific framework.Some easy proofs, which are merely simple applications of definitions, may be sketched oromitted - the detailed writing of these proofs make for excellent learning exercises. Generallyspeaking, readers who wish to have a good understanding of the concepts we introduce shouldrework the course’s proofs by themselves.

Proofs which are considered to be either non-essential to the course or too difficult, shallbe written in small print. These are not studied during the lectures, but are included in thenotes to satisfy the curiosity of motivated students.

The notes are structured as follows: In the first chapter, we show basic results on thetopology of metric spaces, including the notions of complete metric spaces and compact metricspaces. A solid mastery of the contents of this chapter is absolutely imperative.

The second chapter deals with the study of normed vector spaces, fundamental examples ofwhich are function spaces. A key point here is understanding the effects of working in infinitedimension (which is the case in function spaces) on topology. Ascoli’s theorem, a compactnesscriterion for parts of continuous function spaces, illustrates the difficulties which appear ininfinite dimension frameworks.

The third chapter deals with the notion of duality. It may be short, but it is fundamental.Duality is the basis of the theory of distributions, a major breakthrough in analysis at thestart of the XXth century - this will be studied in chapter 8. Beyond the concept of transposesof linear maps, this chapter explains the procedure which allows one to identify the dual ofa Banach space - another Banach space with a weaker notion of convergence, induced by thefact that it is a dual space, that we call “weak-star convergence”.

The fourth chapter is a classic: Hilbert spaces, which extend the notion of Euclidean spacesto infinite dimension.

The fifth chapter studies the spaces of functions which have a finite integral relative to ameasure when elevated to a power p. We start by recalling fundamental results of integrationtheory, without proof. Another important notion, the convolution of functions, is defined,studied, then applied to approximation.

5

The sixth chapter studies the Dirichlet problem on a bounded domain. The goal is to provethat, for any square-integrable function f defined on a bounded, connected open set ⌦ of Rd,there exists a unique solution u in a function space we will define (the Sobolev space H

10 (⌦)),

which solves, in a broader sense than usual, the Laplace equation

�dX

j=1

@2u

@xj2= f,

and such that the function u is zero on the boundary of the open set ⌦. We find this solutionby determining if the infimum of the function

u 7�!Z

⌦

dX

j=1

��@u

@xj

��2dx�

Z

⌦f(x)u(x)dx

on the set of continuously differentiable functions with compact support in ⌦, is reached. Manypreviously established concepts and results come into play here, and ideas from the theory ofdistributions, which will be developed in chapter 8, manifest themselves for the first time.

The seventh chapter deals with the Fourier transform on the space of integrable functionson Rd and many applications. This chapter is fundamental and will be crucial in the followingtwo.

The eighth chapter presents the theory of tempered distributions. We choose not to gointo the general theory of distributions to keep things simple. The basic idea is that, whenwe know how to define an operation on very smooth and rapidly decreasing functions on Rd

(e.g. functions in the Schwartz space), we can use duality to extend it to a space of tempereddistributions, which contain functions (hence “distributions” are also called “generalised func-tions”) as well as some very singular objects. Of course, this chapter contains examples whichmust be studied and understood in order to properly comprehend and apply this theory.

Translator’s note. The convention we use throughout these notes is that “positive” means“strictly positive” (x > 0), and that “increasing” means “strictly increasing” (x < y impliesf(x) < f(y)). For coherence, mathematical notations have been chosen strictly identical tothe original version.

6

Chapter 1

Metric spaces

Introduction

This chapter is a summary of basic results on metric spaces. The first section presents thenotion of metrics, or distances, which is very natural and intuitive, and shows how it allowsone to generalise the convergence of sequences and the continuity of functions (both notionsthat one is already familiar with in the real- and complex-variable settings). Metrics also allowone to give abstract definitions of open and closed subsets, concepts which are constantly usedin functional analysis.

In the second section, we introduce complete metric spaces, which are spaces in which allCauchy sequences are convergent. This notion is fundamental: in these spaces, we can provethat certain sequences converge without needing any prior information about their limit. Acrucial tool, it is very commonly used in analysis to prove existence theorems - the Cauchy-Lipschitz theorem is one such result.

In the third section, we discuss compact metric spaces. Functional analysis requires theuser to go beyond the elementary description of compact subsets of RN .

On the whole, this chapter is rather abstract. Examples, illustrations and applications ofthe fundamental notions presented here will be plentiful in the remainder of the course.

1.1 Definition of metric spaces

Definition 1.1.1. Let X be a set. We call a metric, or distance function, on X any function

d from X ⇥X to R+such that

d(x, y) = 0 () x = y

d(x, y) = d(y, x)

d(x, y) d(x, z) + d(z, y)

The pair (X, d) is called a metric space.

Some examples

• Set X = R and d(x, y) = |x� y|. These define a metric space.

7

• Set X = RN , and choose among the following distance functions:

de(x, y)def=

✓ NX

j=1

(xj � yj)2

◆ 12

d1(x, y)def= max

1jN

|xj � yj |

d1(x, y)def=

NX

j=1

|xj � yj |.

• More generally, let us consider (Xj , dj)1jN , a finite family of metric spaces.

Set X =NY

j=1

Xj , and define

D1

(X ⇥X �! R+

(x, x0) 7�! max1jN

dj(xj , yj) and D1

8><

>:

X ⇥X �! R+

(x, x0) 7�!NX

j=1

dj(xj , yj)(1.1)

Both of these functions are metrics on X.

• Let us return to X = R, let f be an one to one function from R to R, and we define

df (x, y) = |f(x)� f(y)|.

The function df is a metric on X.

The following exercise shows how we can define a metric on the space of sequences ofelements of a given metric space.

Exercise 1.1.1. Let (X, d) be a metric space, and let (an)n2N be a sequence of positive real

numbers such that X

n2Nan < 1.

We consider the set XN

made up of the sequences of elements of X. We define

Da

8<

:

XN ⇥X

N �! R+

(x, y) 7�!X

n2Nmin

�an, d(x(n), y(n))

.

Prove that Da is a metric on XN.

Definition 1.1.2. Let (X, d) be a metric space, x be a point in X and ↵ a positive real

number. The open (resp. closed) ball with centre x and radius ↵, denoted B(x,↵) (resp.

Bf (x,↵)) the set of points y in X such that d(x, y) < ↵ (resp. d(x, y) ↵).

Metrics allow one to give very simple and general definitions for the concepts of the limitof a sequence and the continuity of a function.

Definition 1.1.3 (convergent sequence). Let (xn)n2N be a sequence of elements of a metric

space (X, d) and let ` be a point of X. The sequence (xn)n2N is said to converge to ` if and

only if

8" > 0 , 9 n0 / n � n0 =) d(xn, `) < ".

8

Exercise 1.1.2. We consider a metric space (X, d) and the metric Da defined on XN

in

exercise 1.1.1. Show that a sequence (xp)p2N of elements of XN

converges to x in the sense of

the metric Da if and only if

8n 2 N , limp!1

d(xp(n), x(n)) = 0.

Definition 1.1.4 (continuous function). Let (X, d) and (Y, �) be two metric spaces. We

consider a function f from X to Y and a point x0 in X. The function f is said to be

continuous at the point x0 if and only if

8" > 0 , 9 ↵ > 0 / d(x, x0) < ↵ =) �(f(x), f(x0)) < ".

Proposition 1.1.1 (composition of continuous functions). Let (X, d), (Y, �) and (Z, ⇢) be

three metric spaces, and f et g be two functions, from X to Y and from Y to Z respectively.

Let x0 be a point of X such that f is continuous at x0 and g is continuous at f(x0). Then the

function g � f is continuous at the point x0.

Proof. Since the function g is continuous at f(x0), we have

8" > 0 , 9↵ > 0 / �(y, f(x0)) < ↵ =) ⇢(g(y), (g � f)(x0)) < ".

Since the function f is continuous at x0, we have

9� > 0 / d(x, x0) < � =) �(f(x), f(x0)) < ↵.

As the real number � is chosen for each ", we deduce that

9� > 0 / d(x, x0) < � =) ⇢(g � f(x), (g � f)(x0)) < ".

This ends the proof. ⇤

Proposition 1.1.2 (sequences and continuous functions). Let (X, d) and (Y, �) be two metric

spaces, f be a function from X to Y , and (xn)n2N be a sequence of elements of X. We

assume that (xn)n2N converges to ` and that the function f is continuous at point `. Then

the sequence (f(xn))n2N of elements of Y converges to f(`).

The proof of this proposition is similar to the previous one, and is left as an exercise forthe reader.

Definition 1.1.5 (Interior and closure, open and closed sets). Let (X, d) be a metric space,

and A be a subset of X.

• We call the interior of A, denoted

�

A, the set of points x in X such that there exists an

open ball B(x,↵), with ↵ > 0, which is a subset of A.

• We call the closure of A, denoted A, the set of points x in X such that, for any open

ball B(x,↵), with ↵ > 0, the intersection of B(x,↵) with A is non-empty.

• A set A is said to be dense in X if and only if A = X.

• A set A is said to open if and only if

�

A = A.

• A set A is said to be closed if and only if A = A.

9

Remark The definitions clearly imply that�

A ⇢ A ⇢ A.

Exercise 1.1.3. Let A be a finite part of a metric space (X, d). Prove that A = A.

Proposition 1.1.3. The interior of an open ball B is B. The closure of a closed ball B0is B

0.

Proof. Let y be a point of the open ball B(x,↵), and let us consider the open ball with centre yand radius ↵� d(x, y) (which is a positive number). Using the triangular inequality, we have,for any z in B(y,↵� d(x, y)),

d(x, z) d(x, y) + d(y, z) < d(x, y) + ↵� d(x, y) = ↵.

Now let us consider a point y in the closure of the closed ball with centre x and radius ↵.By definition, for any positive real number �, there exists a point z in B(y,�) \ Bf (x,↵).Once again using the triangular inequality, we get

d(x, y) d(x, z) + d(z, y) < � + ↵.

Hence, for any positive number �, d(x, y) < ↵+ �, so d(x, y) ↵, which ends the proof. ⇤Remarks

• The above proposition states precisely that open balls are open sets, and that closedballs are closed sets (this consistency in terminology is reassuring news).

• The set X is both open and closed. We set the standard that the same occurs for theempty set.

• The proof above shows that the closure of the open ball B(x0,↵) is a subset of the closedball Bf (x0,↵). However, it is not true in general that the closure of an open ball is thecorresponding closed ball. For example, considering a generic set X and the metric d

defined by d(x, y) = 1 if x is not equal to y, and d(x, x) = 0, we have B(x0, 1) = {x0},and this set is also closed, therefore equal to its closure. Meanwhile, Bf (x0, 1) = X.

Proposition 1.1.4. Let A be a subset of a metric space (X, d). We have

(�

A)c = Ac and Ac=

�

(Ac) .

Proof. A point x in X belongs to (�

A)c if and only if

8↵ > 0 , B(x,↵) \Ac 6= ;,

which is means precisely that x belongs to the closure of the complement of A. A point x

belongs to (A)c if and only if9↵ > 0 , B(x,↵) \A = ;,

in other words B(x,↵) ⇢ Ac, which means that x belongs to

�

(Ac). The proposition is proved.⇤Remark Proposition 1.1.4 states that the complement of an open set is closed, and vice-versa.

Proposition 1.1.5. Let A be a subset of a metric space (X, d). A point x in X belongs to A

if and only if there exists a sequence (an)n2N of elements of A such that limn!1

an = x.

10

Proof. Let us assume that x is the limit of a sequence (an)n2N of elements of A. For anypositive real number ↵, there exists an integer n0 such that xn0 belongs to the ball B(x,↵),which implies that B(x,↵) \A 6= ;. So x 2 A.

Conversely, let us assume that x 2 A. Then, for any positive integer n, there exists anelement an of X such that

an 2 A \B(x, n�1)

Consider the sequence (an)n2N that we have defined: it satisfies

d(an, x) 1

n·

The sequence (an)n2N therefore converges to x, and the proposition is proved. ⇤

Proposition 1.1.6. Any union of open sets is open. The intersection of a finite number of

open sets is open. Any intersection of closed sets is closed. The union of a finite number of

closed sets is closed.

Proof. Let (U�)�2⇤ be a family of open sets, and x be a point of U def=[

�2⇤

U�. Let � 2 ⇤ be

such that x 2 U�. Since U� is open (therefore equal to its interior),

9↵ > 0 / B(x,↵) ⇢ U� ⇢ U

so U is open. Now let U =N\

j=1

Uj where each Uj is open. For every j in {1, · · ·N}, there exists

a positive real number ↵j such that B(x,↵j) ⇢ Uj . Set ↵ def= min{↵j , j 2 {1, · · ·N}}. For

every j, the open ball B(x,↵) is a subset of Uj , and is therefore in the intersection U . ⇤Remark For a set X, let us consider ⇥ a part of P(X), which is the set of all subsets of X,which satisfies:

• the empty set and X belong to ⇥,

• if (Uj)1jN is a finite family of elements of ⇥, thenN\

j=1

Uj also belongs to ⇥,

• if (U�)�2⇤ is a family of elements of ⇥, then[

�2⇤

also belongs to ⇥.

This defines what is called a topology on X, with the elements of ⇥ being defined as open setsof X, and closed sets are by definition complements of open sets. As the following propositionwill show, the notions of convergent sequences and continuous functions can be defined interms of open sets.

Proposition 1.1.7. Let (X, d) be a metric space. Let (xn)n2N be a sequence of elements

of X, and x be a point of X. The sequence (xn)n2N converges to x if and only if for any open

subset U of X containing x, there exists an integer n0 such that

8n � n0 , xn 2 U.

Let (Y, �) be another metric space, f be a function from X to Y , and x0 be an element of X.

The function f is continuous at x0 if and only if for any open subset V of Y containing f(x0),there exists an open subset U of X containing x0 such that

f(U) ⇢ V.

11

The proof of this proposition makes for an educational exercise we highly recommend.

Theorem 1.1.1 (Characterisation of continuous functions). Let f be a function between two

metric space (X, d) and (Y, �). The following three statements are equivalent.

• The function f is continuous at every point of X.

• The preimage of an open set is open.

• The preimage of a closed set is closed.

Proof. Assume that f is continuous at every point in X, and let us consider an open subset Vof (Y, �) and a point x in f

�1(V ). Since the set V is open, by definition there exists a positivenumber "0 such that B(f(x), "0) ⇢ V . As the function f is continuous at point x,

9↵ > 0 , f(B(x,↵)) ⇢ B(f(x), "0).

Thus,B(x,↵) ⇢ f

�1⇣f(B(x),↵)

⌘⇢ f

�1(B(f(x), "0)) ⇢ f�1(V ).

Conversely, let x0 be a point of X. For any " > 0, the ball B(f(x0), ") is an open subsetof (Y, �). By assumption, f

�1(B(f(x0), ")) is an open subset of (X, d). So, there exists apositive ↵ such that B(x0,↵) is a subset of f�1(B(f(x0), ")), and therefore

f(B(x0,↵)) ⇢ f

⇣f�1(B(f(x0), "))

⌘⇢ B(f(x0), ").

The equivalence between the second and third statements is a consequence of closed sets beingcomplements of open sets (and vice-versa), and of

f�1(V ) = {x 2 X / f(x) 2 V }

= {x 2 X / f(x) 2 Vc}c

= (f�1(V c))c.

The theorem is proved. ⇤We can now ask questions about the effects of a change of metric on the properties of a

metric space.

Definition 1.1.6. Let X be a set, on which we consider two metrics d and �. We say that

these two metrics are topologically equivalent if and only if the identity map Id is continuous

from (X, d) to (X, �), and from (X, �) to (X, d).

Remark The open sets induced by two topologically equivalent metrics are the same; thusthe same functions are continuous and the same sequences are convergent in both settings.

Proposition 1.1.8. Let X be a set, and d and � be two metrics on X. Then we have the

following property.

The metrics d and � are topologically equivalent if and only if

8x 2 X , 8" > 0 , 9⌘ > 0 / 8y 2 Y ,

d(x, y) < ⌘ =) �(x, y) < " and �(x, y) < ⌘ =) d(x, y) < ".

Proving this proposition merely requires using the definitions; it is left to the reader.

12

Definition 1.1.7. Let A be a subset of a metric space (X, d), and x be a point of X. We call

distance between the point x and the set A the following quantity

d(x,A)def= inf

a2A

d(x, a).

Exercise 1.1.4. Prove that A is the set of points x in X such that d(x,A) = 0.

Proposition 1.1.9. Let A be a subset of a metric space (X, d). The function

dA

⇢X �! R+

x 7�! d(x,A)

is Lipschitz continuous, and 1 is a Lipschitz constant, that is

��d(x,A)� d(x0, A)�� d(x, x0).

Proof. Using the triangular inequality, we have, for any (x, y) in X2 and any point a in A,

d(x, a) d(x, y) + d(y, a).

Since the infimum of a set is a lower bound of this set, we have, for any (x, y) in X2 and any

point a in A,

d(x,A) d(x, y) + d(y, a) and thus d(x,A)� d(x, y) d(y, a).

As the infimum of a set is the largest lower bound of this set, we deduce that

d(x,A)� d(x, y) d(y,A) and thus d(x,A)� d(y,A) d(x, y).

The proof is finished by exchanging the positions of x and y. ⇤We will now introduce the notion of metric subspaces. Let (X, d) be a metric space, and A

be a subset of X. It is natural to consider the metric space (A, d|A⇥A). We have the followingproperty.

Proposition 1.1.10. Let (X, d) be a metric space, and A be a subset of X. Then B, a part

of A, is an open (resp. closed) subset of the metric space (A, d|A⇥A) if and only if there exists

an open (resp. closed) subset eB of X such that B = eB \A.

Proof. We will only deal with open subsets, and the case of closed subsets can be obtainedby taking the complements1 Let eB be an open subset of X, and let us prove that eB \A is anopen set in (A, d|A⇥A). Let a0 be a point in eB \A. Since eB is open in X, there exists an openball (for the metric space (X, d)) such that

B(a0,↵) ⇢ eB.

By intersecting with A, we know that B(a0,↵) \ A is a subset of B. But B(a0,↵) \ A isexactly the set of all a in A such that d(a0, a) < ↵. So B is an open subset of the metricspace (A, d|A⇥A).

Conversely, let B be an open subset of the metric space (A, d|A⇥A). For every a in B, thereexists a positive number ↵a such that

BA(a,↵a) ⇢ A with BA(a,↵a) = {a0 2 A/ d(a, a0) < ↵}.1We strongly recommend writing the details of the closed subset case as an exercise.

13

Set eB def=

[

a2A

BX(a,↵), which is open since it is a union of open sets. Since BA(a,↵) =

BX(a,↵) \A, we have eB \A = B, and the proposition is proved. ⇤To finish off this introduction on metric spaces, we define the notion of diameter in a metric

space.

Definition 1.1.8. Let (X, d) a metric space. We will say that a subset A of X has finite

diameter if and only if there exists a positive number C such that

8(a, a0) 2 A2, d(a, a0) C.

If A has finite diameter, the diameter of A is defined as the supremum of the set of quanti-

ties d(a, a0) with (a, a0) in the set A⇥A.

1.2 Complete spaces

Definition 1.2.1. Let (X, d) be a metric space. We call a Cauchy sequence in X any se-

quence (xn)n2N of elements of X that satisfies

8" > 0 , 9n0 2 N / 8n � n0 , 8m � n0 , d(xn, xm) < ".

We note that if (xn)n2N is a sequence of elements of a metric space (X, d) that is convergentto a point ` in X, then, since

d(xn, xm) d(xn, `) + d(`, xm),

it is also a Cauchy sequence. Complete spaces are metric spaces in which the converse is true.This is told precisely in the following definition.

Definition 1.2.2. Let (X, d) be a metric space. This space is said to be complete if and only

if all Cauchy sequences in X are convergent.

Let us provide a few examples of complete spaces. A fundamental example is the space Rendowed with the metric d(x, y) = |x� y|, which is complete due to the way it is constructed.We will now show how to make complete spaces based on some we already know. In otherwords, we will look for operations on metric spaces for which the completeness of these spacesis preserved.

Proposition 1.2.1. Let (X1, d1), · · · , (XN , dN ) be a family of N complete metric spaces. If

we set

X = X1 ⇥ · · ·⇥XN and d((x1, · · · , xN ), (y1, · · · , yN )) = max1jN

dj(xj , yj),

then the metric space (X, d) is complete.

The proof is left as an exercise. An immediate consequence is that the space RN , endowedwith any one of the metrics de, d1 or d1, is complete.

The following exercise provides us with an interesting example.

Exercise 1.2.1. Let (X, d) be a complete metric space. We consider on XN

the metric Da

introduced in exercise 1.1.1. Show that the metric space (XN, Da) is complete.

14

Proposition 1.2.2. Let (X, d) be a complete space. We consider a subset A of X. The metric

space (A, d|A⇥A) is complete if and only if A is closed.

Proof. Assume that (A, dA⇥A) is complete. Let (an)n2N be a sequence of elements of A thatconverges to x in X. The sequence (an)n2N is a Cauchy sequence in X made up of elementsof A, so it is a Cauchy sequence in the space (A, d|A⇥A), which is complete. Therefore, thereexists a in A such that the sequence (an)n2N converges to a in (A, d|A⇥A), which is a metricsubspace of (X, d). Since the limit of a sequence is unique, we have a = x and x is in A.

Conversely, let us assume that A is closed, and consider a Cauchy sequence (an)n2N in themetric space (A, d|A⇥A). It is therefore a Cauchy sequence in (X, d) which is complete, so itis convergent to x in X. The fact that A is closed implies that x is in A, and thus, (A, d|A⇥A)is complete. ⇤

When a metric space (X, d) is complete, we can prove the existence of some objects. Thefollowing theorem is the most spectacular illustration of this.

Theorem 1.2.1 (contracting map fixed-point theorem, Picard). Let f be a function from a

complete metric space (X, d) to itself such that there exists a number k in ]0, 1[ such that, for

all (x, y) 2 X2,

d(f(x), f(y)) kd(x, y).

Then there exists a unique fixed point z in X, satisfying f(z) = z.

Proof. For a given an element x0 in X, we consider the sequence (xn)n2N defined by theiterative formula xn+1 = f(xn). We therefore write that

d(xn+1, xn) = d(f(xn), f(xn�1))

kd(xn, xn�1).

We iterate these multiplications, and get that

d(xn+1, xn) knd(x1, x0).

Thus, for any pair of integers (n, p), we have

d(xn+p, xn) pX

m=1

d(xn+m, xn+m�1)

d(x1, x0)pX

m=1

kn+m�1

kn

1� kd(x1, x0).

Hence, the sequence (xn)n2N is a Cauchy sequence. Let z be its limit. As the function f iscontinuous, seeing that it is Lipschitz continuous, we take the limit in the iterative formuladefining (xn)n2N, and get that z = f(z).

It remains to prove that the fixed point is unique. Let z1 and z2 be two solutions ofz = f(z). The hypothesis on f leads to

d(z1, z2) kd(z1, z2).

The fact that k must be smaller than 1 means that d(z1, z2) = 0, hence z1 = z2. Thus, thetheorem is proved. ⇤

15

Remark. This theorem is the key ingredient for numerous existence and uniqueness the-orems. An important example is the Cauchy-Lipschitz theorem, guaranteeing existence anduniqueness for solutions of ordinary differential equations.

We shall now prove Baire’s classic theorem, which has many applications in functionalanalysis. There will be some examples of these in chapter 2.

Theorem 1.2.2 (Baire category theorem). Let (X, d) be a complete metric space. We consider

a sequence (Un)n2N of dense open subsets of X. Then,

\

n2NUn is also dense.

Proof. The proof relies largely on the following lemma, which is interesting in itself.

Lemma 1.2.1. Let (X, d) be a complete metric space, and (Fn)n2N be a non-increasing

sequence of non-empty closed subsets of X, such that the diameter of Fn tends to zero. Then

there exists an element x in X such that

\

n2NFn = {x}.

Proof. Let us consider a sequence (xn)n2N of elements of X such that, for every integer n,xn belongs to Fn. Since the sequence of sets (Fn)n2N is non-increasing (in the sense thatFn+1 ⇢ Fn), we have

8p 2 N , d(xn, xn+p) �(Fn).

The fact that the diameter of the sets Fn tends to zero implies that the sequence (xn)n2Nis a Cauchy sequence, and therefore it converges to a point x in X. By making p go toinfinity in the above statement, we get that every sequence (xn)n2N such that xn is in Fn

satisfies d(xn, x) �(Fn). This ends the proof of the lemma. ⇤Back to the proof of theorem 1.2.2. Let V be an open subset of X. We are going to provethat

\

n

Un \ V 6= ;, which is enough to prove the theorem.

The open set U0 is dense in X, so U0 \ V is a non-empty open set. Hence, there exist apositive number ↵0 (which we can assume is less than 1) and a point x0 in X such that

Bf (x0,↵0) ⇢ U0 \ V. (1.2)

The open set U1 is also dense, so the set U1 \ B(x0,↵0) is a non-empty open set. So thereexist a positive number ↵1 (which we can assume is less than 1/2) and a point x1 in X suchthat

Bf (x1,↵1) ⇢ U1 \B(x0,↵0).

We proceed by induction and assume that we have built a sequence (xj)0jn of elementsof X and a sequence (↵j)0jn such that, for each j n, we have

↵j 1

j + 1and Bf (xj ,↵j) ⇢ Uj \B(xj�1,↵j�1). (1.3)

The open set Un+1 is dense, so the set Un+1 \ B(xn,↵n) is a non-empty open set. Thereforethere exist a positive number ↵n+1 (which we can assume is less than 1/(n + 2)) and apoint xn+1 in X such that

Bf (xn+1,↵n+1) ⇢ B(xn,↵n) \ Un+1.

16

Applying lemma 1.2.1 to the sequence Fn = Bf (xn,↵n) implies that there exists a point x thatbelongs to the intersection of the closed balls Bf (xn,↵n). Given the equations (1.2) and (1.3),we have

8n 2 N , x 2 V \\

jn

Uj .

The Baire category theorem is proved. ⇤Often, the following statement is used, which is the Baire theorem in which we consider

the complements.

Theorem 1.2.3. Let (X, d) be a complete metric space, and consider a sequence (Fn)n2N of

closed subsets of X which have empty interiors. Then the interior of the set

[

n2NFn is empty.

The Baire category theorem is also used as formulated in the following corollary, the proofof which is very easy, and is left as an exercise.

Corollary 1.2.1. Let (Fn)n2N be a sequence of closed sets in a complete metric space (X, d),

such that the union of all the Fn is X. Then, there exists an integer n0 such that

�

Fn0 6= ;.

In other words, a complete metric space cannot be the union of countably many closed setswith empty interiors. One noteworthy non-complete metric space is built this way: Q equippedwith the distance d(x, y) = [x� y|.

Let us now introduce the notion of uniformly continuous functions.

Definition 1.2.3. Let (X, d) and (Y, �) be two metric spaces, and we consider a function f

from X to Y . The function f is said to be uniformly continuous if and only if

8" > 0 , 9↵ > 0 / d(x, x0) < ↵ =) �(f(x), f(x0)) < ".

Here are some examples and counter-examples. The function x 7�! x2 is uniformly con-

tinuous on any bounded interval [a, b], but not on R. The function x 7�!p

|x| is uniformlycontinuous on R.

Theorem 1.2.4 (extension of uniformly continuous functions). Let (X, d) and (Y, �) be two

metric spaces, A be a dense subset of X, and f be a uniformly continuous function from (A, d)to (Y, �). If Y is complete, then there exists a unique uniformly continuous function effrom (X, d) to (Y, �) such that ef|A = f .

Proof. Let us consider an element x in X, and we will try to give a value to ef(x). Since theset A is dense in X, proposition 1.1.5 ensures that there exists a sequence (an)n2N of elementsof A which converges to x. Also, the function f is uniformly continuous on A, which meansthat

8" > 0 , 9↵ / 8(a, b) 2 A2, d(a, b) < ↵) �(f(a), f(b)) < ".

As the sequence (an)n2N converges, it is a Cauchy sequence, so there exists an integer n0 suchthat

8n � n0 , 8p , d(an, an+p) < ↵.

This means that the sequence (f(an))n2N is a Cauchy sequence in Y , so it converges to alimit y.

The first thing to check is that this limit does not depend on the sequence (an)n2N wechose to construct it. To see that this is true, take (an)n2N and (bn)n2N, two sequences that

17

converge to x. Reasoning similarly to above, the uniform continuity of the function f on A

gives uslimn!1

�(f(an), f(bn)) = 0.

So the limit y does indeed not depend on the choice of the sequence (an)n2N. Therefore, wedefine ef as follows:

ef(x) = limn!1

f(an) for any sequence (an)n2N 2 AN/ lim

n!1an = x.

Since the function f is continuous on A, it is clear that ef|A = f . We must now prove that efis uniformly continuous on X. The uniform continuity of ef can be written mathematically as

8" > 0 , 9↵ > 0 / 8(a, b) 2 A2, d(a, b) < ↵) �(f(a), f(b)) < ". (1.4)

So let us consider a pair (x, y) of elements of X such that d(x, y) < ↵. There exist twosequences (an)n2N and (bn)n2N of elements of A such that

limn!1

an = x and limn!1

bn = y.

There must exist an integer n0 such that

n � n0 =) d(an, bn) < ↵.

Therefore, using equation (1.4), we have

�(f(an), f(bn)) < ".

Taking the limit, we obtain

d(x, y) < ↵ =) �( ef(x), ef(y)) ",

which ends the proof of the theorem. ⇤

1.3 Compactness

An important concept we will use in this section is that of cluster points of a sequence.

Definition 1.3.1. Let (xn)n2N be a sequence of elements of a metric space (X, d). We call the

set of cluster points of the sequence (xn)n2N, and denote it Adh(xn) (from the French “valeur

d’adhérence”), the set -which may be empty- defined by

Adh(xn)def=\

n2NAn with An

def= {xm ,m � n}.

Examples In R endowed with the metric |x� y|, the sequence (xn)n2N defined by xn = n issuch that Adh(xn) = ;, and the sequence (yn)n2N defined by yn = (�1)n+1 satisfies Adh(yn) ={�1, 1}.

Proposition 1.3.1. Let (xn)n2N be a sequence in a metric space (X, d) that we assume

converges to a point ` in X. Then Adh(xn) = {`}.

Proof. By definition of the limit of a sequence, for every " > 0, there exists n0 such that theset An0 defined above is included in the open ball B(`, "). So, if x is a point of X not equalto `, by taking " smaller than d(`, x), we see that x 62 An0 , thus x 62 An for all n � n0. Hencethe proposition is proved. ⇤

18

Remark It is possible that a sequence (xn)n2N satisfies Adh(xn) = {`}, but does not converge.Consider, for example, the sequence of real numbers defined by

x2n = 2n and x2n+1 =1

2n+ 1·

Even though this sequence does not converge to 0, we still have Adh(xn) = {0}.

Proposition 1.3.2. In a metric space (X, d), any Cauchy sequence that possesses a cluster

point ` converges to `.

Proof. Let us consider a Cauchy sequence (xn)n2N with a cluster point `. As it is a Cauchysequence, there exists an integer n0 such that

8n � n0 , 8m � n0 , d(xn, xm) "

2·

As the sequence has a cluster point `, there exists an integer n1 � n0 such that

d(xn1 , `) <"

2·

We therefore see that, for every integer n � n0, we have

d(xn, `) d(xn, xn1) + d(xn1 , `) < ".

The proposition is proved. ⇤Proposition 1.3.3. Let (xn)n2N be a sequence of elements of a metric space (X, d). A point `

in X belongs to Adh(xn) if and only if there exists an increasing function � from N to N such

that

limn!1

x�(n) = `.

Proof. Before we starting the proof of this proposition, we remark that an increasing functionfrom N to N satisfies

8n 2 N , �(n) � n. (1.5)

This property is proved by induction. The statement is naturally true for n = 0. Now, weassume that it is true for some n. Since f is increasing, we have �(n+ 1) > �(n) � n, whichimplies that �(n + 1) � n + 1. In particular, the sequence (�(n))n2N must tend to infinitywhen n does so.

We return proving the proposition. If ` belongs to Adh(xn), we define the function � byinduction as follows: we choose �(0) = 0, then get �(n+ 1) using �(n) like so

�(n+ 1)def= min

nm > �(n) /d(xm, `) <

1

n+ 1

o·

The function � is increasing by construction, and we have, for every n � 1,

d(x�(n), `) 1

n, so lim

n!1x�(n) = `.

Conversely, if there exists an increasing function � such that limn!1

x�(n) = `, then for every " >0, there exists n0 such that

8n � n0 , d(x�(n), `) < ".

Using (1.5), that is �(n) � n, we deduce that

8" > 0 , 8n 2 N , 9m � n / d(xm, `) < ",

which means that ` 2 Adh(xn), hence the theorem is proved. ⇤

19

Definition 1.3.2. We call extraction function, any increasing function from N to N. If (xn)n2Nis a sequence of elements of a set X, a subsequence is a sequence which can be written

as (x�(n))n2N, where � is an extraction function.

The notion of compact space can be seen from two angles: one using sequences, the otherwith balls that cover the space. The equivalence of these two viewpoints is given by thefollowing theorem.

Theorem 1.3.1. Let (X, d) be a metric space. The following two statements are equivalent:

i) every sequence of elements of (X, d) has a cluster point;

ii) the metric space (X, d) is complete, and we have

8" > 0 , 9(xj)1jN /X =N[

j=1

B(xj , "). (1.6)

Definition 1.3.3. Let (X, d) be a metric space. We say that X is a compact space if and

only if one of the two conditions of the above theorem is satisfied.

Let us start a list of examples of compact metric spaces by noting that any finite metricspace is compact.

Theorem 1.3.2. The metric spaces ([a, b], |x� y|) are compact.

Proof. If we use condition ii), which is reasonable because the space ([a, b], |x�y|) is complete,it suffices to see that, for every positive ",

[a, b] = [a, a+ "[[✓N"�1[

k=1

]a+ (k � 1)", a+ (k + 1)"[

◆[]b�N"", b] with N"

def=

b� a

"

�·

Using criterion i) requires a bit more work. Let (xn)n2N be a sequence in the interval [a, b],and we consider the set An

def= {xm , m � n}, which is a set in R which has an upper bound.

It therefore has a supremum, which we denote Mn. The sequence (Mn)n2N is non-increasingand bounded from below, so it converges. The reader will notice that the limit of the sequence(Mn)n2N is a cluster point of (xn)n2N (it is, in fact, the largest cluster point). ⇤Proof of theorem 1.3.1. By using proposition 1.3.2, in order to prove that i) implies ii),it suffices to show that i) implies statement (1.6). We will prove the contrapositive. Letus assume that the statement (1.6) is not satisfied. This means that there exists a positivenumber ↵ such that we cannot cover X with a finite number of balls with radius ↵. Let x0 beany element of X. There exists an element x1 in X which does not belong to the ball B(x0,↵).Now let (x0, · · · , xp) be a p-tuple of elements of X such that, for every m not equal to n, wehave d(xm, xn) � ↵. We assume that

p[

n=1

B(xn,↵) 6= X.

So, there exists a point xp+1 in X that does not belong to the above union of balls. Byinduction, we construct a sequence (xn)n2N such that

m 6= n ) d(xm, xn) � ↵.

20

Such a sequence cannot have a cluster point, so we have proved that i) implies ii).Let us now assume ii), and consider an infinite sequence (xn)n2N of elements of X. The

covering hypothesis implies that there exists an element ↵0 in X such that the set X0 definedby

X0def=nm/ xm 2 F0

def= Bf

⇣↵0,

1

2

⌘o

is infinite. We set�(0)

def= minX0.

Likewise, there exists an element ↵1 in X such that the set X1 defined by

X1def=nm 2 X0 / xm 2 F1

def= F0 \Bf

⇣↵1,

1

4

⌘o

is infinite. We then set�(1)

def= min

�m 2 X1 /

pm > �(0)

.

Let us assume that we have built a sequence (Fm)1mn of closed subsets of X, an increasingsequence of integers (�m)1mn such that �(Fm) 2�m, and a sequence (Xm)1mn of subsetsof N defined such that

Xm

def= {m0 2 Xm�1/ xm0 2 Fm�1}

is infinite. Then there exists an element ↵n+1 in X such that

Xn+1def=nm/ xm 2 Fn+1

def= Fn \Bf

⇣↵n+1,

1

2n+1

⌘o

is infinite. We set�(n+ 1)

def= min{m 2 Xn+1 /

pm > �(n)}.

We therefore have, for every positive integer n,

x�(n) 2 Fn and �(Fn) 1

2n·

By lemma 1.2.1, there exists a point ` which belongs to the intersection of all the Fn. Bydefinition of the diameter of a set, we have

8n 2 N , d(x, x�(n)) 1

2n·

This ends the proof of the theorem. ⇤The following proposition yields many examples of compact spaces.

Proposition 1.3.4. Let (X1, d1), · · · , (XN , dN ) be a family of N compact metric spaces. Set


dj(xj , yj).

The metric space (X, d) is compact.

Proof. Let (xn)n2N be a sequence of elements in X. By definition, for each integer j, thereexists a sequence (xjn)n2N in Xj such that xn = (x1n, · · ·xNn ). By definition of a compact space,there exists an extraction function �1 and a point x

1 in X1 such that

limx1�1(n)

= x1.

21

Likewise, there exists an extraction function �2 and a point x2 in X2 such that

limx2�1��2(n)

= x2.

We reiterate this process N times, and we construct an extraction function � by setting

�def= �1 � · · · � �N ,

so that, for every j in {1, · · · , N}, we have

limn!1

xj

�(n) = xj.

Using the definition on the distance we defined on X, we see that the sequence x�(n) convergesto (x1, · · · , xN ). The result is proved. ⇤

Definition 1.3.4. Let (X, d) be a metric space. A subset A is said to be a compact subset

of X if and only if the metric space (A, d|A⇥A) is compact.

Proposition 1.3.5. Let A be a compact subset of a metric space (X, d). If A is compact,

then A is closed and has finite diameter, which means that

�(A)def= sup

(a,a0)2A2d(a, a0) < 1.

Proof. Let x be a point in A. By proposition 1.1.5, there exists a sequence (an)n2N of elementsin A which converges to x. As we assume that A is compact, there exists an extractionfunction � and a point ` in A such that lim

n!1a�(n) = `. Since the sequence (a�(n))n2N also

converges to x, the uniqueness of the limit of a sequence implies that ` = x, and thereforethat x 2 A. So A is a closed subset of X.

Now we prove that a compact subset must have finite diameter. Let us consider twosequences (an)n2N and (bn)n2N of elements in A such that

limn!1

d(an, bn) = �(A).

Since A is compact, there exists an extraction function �0 and a point a in A such that thesequence (a�0(n))n2N converges to a. Likewise, there exists an extraction function �1 and apoint b in A such that the sequence (b�1(n))n2N converges to b. Taking the limit in the aboveequation, we get

�(A) = limn!1

d(a�0��1(n), b�0��1(n)) = d(a, b).

The diameter of A is therefore finite. ⇤

Proposition 1.3.6. Let A be a subset of a complete metric space (X, d). Its closure A

is compact if and only if for every positive real number ", there exists a finite number of

points (xj)1jN in X such that

A ⇢N[

j=1

B(xj , ") .

Proof. If A is compact, then we are just rewriting definition 1.3.3. Conversely, let us assumethe covering property for a subset A of X. First, we prove that the property is also true forits closure A.

22

We are assuming that, for any positive number ", there exists a finite sequence (xj)1jN

of elements in X such that

A ⇢N[

j=1

B

⇣xj ,

"

2

⌘·

Let x be a point in A. By definition, there exists a point a in A such that d(x, a) is smallerthan "/2. So there exists an integer j in {1, · · · , N} such that a belongs to B(xj , "/2). Thus,

d(x, xj) d(x, a) + d(a, xj) < ",

so x belongs to B(xj , "), which shows that the covering property is satisfied by A. As themetric space (X, d) is complete, in order to prove that A is compact, it suffices to show thatwe have the covering property with the points xj being in the set A. We know that, for everypositive number ", there exists a finite sequence (xj)1jN of elements of X such that

A ⇢N[

j=1

B

⇣xj ,

"

2

⌘·

Of course, we only consider points xj such that the open ball with centre xj and radius "/2has a non-empty intersection with A. Let aj be a point in said intersection. Then, for any a

in A, there exists j such that

d(a, aj) d(a, xj) + d(xj , aj) < ",

which ends the proof. ⇤From theorem 1.3.2 and propositions 1.3.4 and 1.3.5, we deduce the following corollary.

Corollary 1.3.1. A closed subset A in Rdis compact if and only if

9r > 0 / A ⇢ [�r, r]d.

Exercise 1.3.1. Let A be a subset of a metric space (X, d). Prove that the closed set A is

compact if and only if any sequence of elements of A has a cluster point in A.

Proposition 1.3.7. Let A be a compact subset of a metric space (X, d). If B is a closed

subset of X that is contained in A, then B is also compact.

Proof. Let (bn)n2N be a sequence of elements of B. Since B is a subset of A, which is compact,there exists an extraction function � such that (b�(n))n2N converges to `, which is an elementof A. Since B is closed in X, ` 2 B = B. Hence, B is compact. ⇤

Theorem 1.3.3 (Heine). Let (X, d) and (Y, �) be two metric spaces, and let f be a continuous

map from X to Y . Then, for any compact subset A of X, f(A) is a compact subset of Y .

Moreover, if X is compact, then f is uniformly continuous from X to Y .

Proof. Let (yn)n2N be a sequence of elements in f(A), that is, there exists a sequence (an)n2Nof elements in A such that yn = f(an). Since A is a compact subset of X, there exists apoint a

0 in A and an extraction function � such that

limn!1

a�(n) = a0

23

As the function f is continuous, we have

limn!1

y�(n) = f(a0) 2 f(A).

This proves the first part of the theorem.To prove the second part, we consider the contrapositive. We assume that the metric

space (X, d) is compact, and consider a function f on X which is not uniformly continuous.This means that there exists a positive real number " and two sequences (xn)n2N and (yn)n2Nof elements in X such that

8n 2 N , d(xn, yn) 1

n+ 1and �(f(xn, yn) � "0.

As the metric space (X, d) is compact, there also exists an extraction function � and a point xin X such that lim

n!1x�(n) converges to x. As the distance between xn and yn converges to

zero, the sequence (y�(n))n2N also converges to x. The triangular inequality leads to

�(f(x�(n)), f(x)) + �(f(x), f(y�(n)) � �(f(x�(n), f(y�(n)) � "0,

which implies that the function f is not continuous at the point x, by proposition 1.1.2. Thetheorem is proved. ⇤

Corollary 1.3.2. Let f be a continuous function from (X, d) to (R, |x � y|). If (X, d) is

compact, then the function f has a minimum and a maximum, i. e. there exist xm and xM

in X such that

8x 2 X , f(xm) f(x) f(xM ).

Proof. By Heine’s theorem, theorem 1.3.3, the set f(X) is compact in R, so it is closed anda subset of some bounded interval [a, b]. This set therefore has an infimum and a supremum,which are cluster points of the closed set f(X), and therefore, they belong to f(X). ⇤

We will now give a characterisation of compactness in terms of covers by open sets.

Theorem 1.3.4. Let (X, d) be a metric space. The following three statements are equivalent.

i) For any family (U�)�2� of open subsets of X that cover X, which means

X =[

�2⇤

U�,

we can extract a finite cover of X, i. e. there exists a finite sequence (�j)1jN such that

X =N[

j=1

U�j .

ii) For any family (F�)�2� of closed subsets of X, we have

8N , 8(�1 · · · ,�N ) 2 ⇤N/

N\

j=1

F�j 6= ; =)\

�2⇤

F� 6= ;.

iii) Every sequence of elements of X has a cluster point.

24

Proof. The equivalence of parts i) and ii) can be proved by considering the contrapositives. Indeed,the contrapositive of ii) is equivalent to:

“For any family (F�)�2⇤ of closed subsets of X,

\

�2⇤

F� = ; =) 9(�1 · · · ,�N ) 2 ⇤N/

N\

j=1

F�j = ;.00

This is seen to be equivalent to i) by taking the complements of the closed sets F�.Let us now prove that ii) implies iii). By definition of the set of cluster points, we have

Adh(xn) =\

n

An with An

def= {xm ,m � n}.

The sequence (An)n2N is a non-increasing sequence of non-empty closed sets. So, by property ii), theintersection of all the An (which is the set of cluster points) is non-empty, which proves point iii).

Let us now prove that iii) implies i), which is the trickiest part of the proof. We have to deducea property about covers by open sets from a property on sequences. The following lemma is a crucialelement, and it can be useful to prove other properties on compact spaces.

Lemma 1.3.1 (Lebesgue). Let (X, d) be a compact metric space. For any family (U�)�2⇤ of opensets which cover X, there exists a positive real number ↵ such that

8x 2 X , 9� 2 ⇤ / B(x,↵) ⇢ U�.

Proof. To prove this lemma, let us consider the function � defined by

�

⇢X �! R?

+

x 7�! sup {� / 9� / B(x,�) ⇢ U�}.

Proving that the function � has a minimum suffices to get the result. To prove this, let us establishthe following.

8" > 0 , 8x 2 X , 9� / d(x, y) < � =) �(y) > �(x)� ". (1.7)

Let x be a point in X, and " be a positive number smaller than �(x). Assume that y is a point in X

such that d(x, y) < "/2. The triangular inequality implies that

B(y, �(x)� ") ⇢ B(x, �(x)� "/2) ⇢ U�,

so the statement (1.7) is proved.Now, let us consider a sequence (xn)n2N of elements in X such that lim

n!1�(xn) = inf

x2X

�(x). Byassumption, we can extract from the sequence (xn)n2N a subsequence that converges to a point x1in X. We will denote this subsequence (xn)n2N. By statement (1.7), for any " > 0, there exists aninteger n0 such that

n � n0 =) �(xn) > �(x1)� ".

Taking the limit, we deduce thatinfx2X

�(x) � �(x1)� ",

which is valid for any positive real number ". By definition of the infimum, we get that �(x1) =infx2X

�(x), and therefore infx2X

�(x) is positive. This ends the proof of Lebesgue’s lemma. ⇤

Conclusion of the proof of theorem 1.3.4To prove that iii) implies i), we assume that X is compact (that is, any sequence has a converging

subsequence), and we consider a family (U�)�2⇤ of open subsets of X that cover X. The aim is toextract a finite ‘subcover’. Let ↵ > 0 be the radius given by Lebesgue’s lemma. By definition 1.3.3,there exists a finite number of balls B1, . . . , BN with radius ↵ that cover X. Given the property that ↵

25

has, owing to Lebesgue’s lemma, each of these balls is included in some open set U�j in the family,and therefore

X ⇢N[

j=1

Bj ⇢N[

j=1

U�j .

This ends the proof of the theorem. ⇤

Exercise 1.3.2. Let (X, d) be a metric space. We consider the set XN

of sequences of elements

in X, and the metric Da on XN

given in exercise 1.1.1.

1) Prove that the sequence (xp)p2N is a Cauchy sequence in XN

if and only if, for every

integer n, the sequence (xp(n))p2N is a Cauchy sequence in X.

2) Deduce that if the metric space (X, d) is complete, then (XN, Da) is also complete.

3) Prove that if the metric space (X, d) is compact, then (XN, Da) is also compact.

4) Prove that U is an open subset of (XN, Da) if and only if for every x 2 U , there exists

a finite subset J of N and a positive number ↵ such that

8y 2 XN, 8j 2 J , d(y(j), x(j)) < ↵ =) y 2 U.

26

Chapter 2

Normed spaces, Banach spaces

This chapter is devoted to the study of Banach spaces, which are normed vector spaces thatare complete with respect to the metric associated with the norm. These are the basic objectsof functional analysis. This chapter provides our first opportunity to use the fundamentalconcepts of metric topology we saw in the previous chapter on concrete examples.

In the first section, the notion of norm and metric associated with a norm is introduced,and, beyond RN , we present and study spaces of bounded continuous functions and spacesof sequences which are summable when elevated to a power p. These are elementary andimportant examples that the reader should keep in mind as basic models of Banach spaces.

In the second section, we study the space of linear continuous maps between Banach spaces.It is important for the rest of the course to fully assimilate the continuity criterion for linearmaps, as well as the notion of norm of a linear map. This part ends with a continuity criterionfor multilinear maps.

In the third section, we study the case of vector spaces with finite dimension N , and showthat all of these can be identified with RN .

The fourth section deals with Ascoli’s theorem, which provides us with a compactnesscriterion for subsets of the space of continuous functions from a compact space to a Banachspace. This theorem allows one to witness how far from true the equivalence between beingcompact and being closed and bounded is in general, unlike in the finite dimension case.

The fifth part contains results on dense sets in the space of continuous functions from acompact set to R or C. We take the opportunity to introduce the concept of separability,which will be put to use in the next chapter.

Finally, we point out that, for the rest of the course, K will denote R or C.

2.1 Definitions of normed spaces and Banach spaces

Definition 2.1.1. Let E be a vector space on K. A map N from E to R+is called a semi-norm

on E if and only if the two following conditions are verified.

• N(x+ y) N(x) +N(y),

• For any � in K, we have N(�x) = |�|N(x).

A map N from E to R+is a norm on E if and only if it is a semi-norm with the property:

N(x) = 0 implies x = 0.

27

Definition 2.1.2. Let E be a vector space and N be a norm on E. The pair (E,N) is called

a normed space.

Definition 2.1.3. Let N1 and N2 be two norms on a vector space E. These two norms are

said to be equivalent if and only if there exists a constant C such that

8x 2 E , C�1

N1(x) N2(x) CN1(x).

Notation. Most often, a norm is denoted k · kE , or k · k.

Proposition 2.1.1. Let (E, k · k) be a normed space. Then, the map defined as follows,

⇢E ⇥ E �! R+

(x, y) 7�! kx� yk

is a metric on E; it is called the metric induced by the norm k · k.

The easy proof of this is left as an exercise.

Convention. Unless it is expressly mentioned otherwise, we will always consider the space E

to be endowed with the metric structure we have just described.

Definition 2.1.4. Let (E, k · k) be a normed space. We say that (E, k · k) is a Banach space

if and only if the metric space (E, d), in which d is the metric induced by the norm k · k(i.e. d(x, y) = kx� yk), is a complete space.

We will now give a sequence of examples of Banach spaces.

Proposition 2.1.2. For p in the interval [1,1[, we consider the map defined by the following:

k · k`p

8><

>:

KN �! R+

(xj)1jN 7�!⇣ NX

j=1

|xj |p⌘ 1

p.

This is a norm on KN, and (KN

, k · k`p) is a Banach space. Moreover, for any pair (p, q)in [1,1[2 such that p q, we have

kxk`1def= sup

1jN

|xj | kxk`q kxk`p N1� 1

q kxk`q N sup1jN

kxk`1 . (2.1)

Proof. We begin by proving what is known as Hölder’s inequality. For any p in the inter-val [1,1[, we have, for any (a, b) in KN ⇥KN ,

��NX

j=1

ajbj

�� kak`pkbk`p0 with1

p+

1

p0= 1. (2.2)

To prove this inequality (which is obvious when p = 1 or p = 1), we note that the concavenature of the logarithm function implies that

8(a, b) 2]0,1[2 , 8✓ 2 [0, 1] , ✓ log a+ (1� ✓) log b log(✓a+ (1� ✓)b).

As the exponential function is increasing, we deduce that

a✓b1�✓ ✓a+ (1� ✓)b,

28

which is often stated, equivalently, as: for any ✓ in the interval ]0, 1[,

ab ✓a1✓ + (1� ✓)b

11�✓ . (2.3)

Dividing the vectors by their norms, we can assume that kak`p = kbk`p

0 = 1. Then,

��NX

j=1

ajbj

�� NX

j=1

|aj | |bj |

1

p

NX

j=1

|aj |p +⇣1� 1

p

⌘|bj |p

0

1.

Now that inequality (2.2) is proved, we can write that

NX

j=1

|xj + yj |p NX

j=1

|xj | |xj + yj |p�1 +NX

j=1

+|yj | |xj + yj |p�1

✓⇣ NX

j=1

|xj |p⌘ 1

p+⇣ NX

j=1

|yj |p⌘ 1

p

◆✓ NX

j=1

|xj + yj |(p�1) pp�1

◆1� 1p

✓⇣ NX

j=1

|xj |p⌘ 1

p+⇣ NX

j=1

|yj |p⌘ 1

p

◆✓ NX

j=1

|xj + yj |p◆1� 1

p

.

Simplifying this, we get that, for p in the interval ]1,1[, the map k · k`p is a norm (thecases p = 1 and p = 1 are again obvious).

Now, we prove (2.1). For any (p, q) in [1,1]2 such that p < q, we have

xj

kxk`q 1.

This implies that ��xj

kxk`q

��q

��

xj

kxk`q

��p

.

Summing all the terms, we get

1 ⇣kxk`pkxk`q

⌘p

which proves the result. ⇤

Proposition 2.1.3. Let X be a set, and (E, k · k) be a Banach space. We consider B(X,E),the set of bounded functions from X to E, i.e. functions f that satisfy

kfkB(X,E)def= sup

x2X

kf(x)k < 1.

Then (B(X,E), k · kB(X,E)) is a Banach space. Moreover, if X is endowed with a metric d, we

can define Cb(X,E), the set of continuous functions in B(X,E). Then (Cb(X,E), k · kB(X,E))is a Banach space.

29

Proof. Let (fn)n2N be a Cauchy sequence in B(X,E). By definition, we have

8" , 9n0 / 8n � n0 , 8p , 8x 2 X , kfn(x)� fn+p(x)k < ". (2.4)

In particular, for every x, the sequence (fn(x))n2N is a Cauchy sequence in E. As this spaceis complete, the sequence converges. So, for every x in X, there exists an element in E, whichwe denote f(x), such that

limn!1

fn(x) = f(x).

We must check that f belongs to B(X,E). By inequality (2.4), in which we take " = 1, wehave

8p , 8x 2 X , kfn0(x)� fn0+p(x)k < 1.

Taking the limit as p tends to infinity, we get

8x 2 X , kfn0(x)� f(x)k 1.

As the function fn0 is bounded, we see that f is also bounded. Indeed, we have

kf(x)� f(x0)k kf(x)� fn0(x)k+ kfn0(x)� fn0(x0)k+ kfn0(x

0)� f(x0)k 2 + kfn0(x)� fn0(x

0)k.

Now, we must check that the sequence (fn)n2N converges to f in the space B(X,E). In orderto do so, we take the limit as p tends to infinity in inequality (2.4), which yields

8" > 0 , 9n0 2 N / 8n � n0 , 8x 2 X , kfn(x)� f(x)k ",

which ends the proof that B(X,E) is complete.To prove that Cb(X,E) is complete, it suffices, by proposition 1.2.2, to prove that Cb(X,E)

is closed in B(X,E). In other words, we must prove that a uniform limit of continuousfunctions is continuous. For this, we consider a sequence (fn)n2N of elements of Cb(X,E)which converges to f in B(X,E), and we must show that f is continuous from X to E. Bydefinition, for any " > 0, there exists an integer n0 such that

8x 2 X , kfn0(x)� f(x)k <"

4· (2.5)

Repeatedly using the triangular inequality and (2.5) above, we can write, for any pair (x0, x)in X ⇥X,

kf(x)� f(x0)k kf(x)� fn0(x)k+ kfn0(x)� fn0(x0)k+ kfn0(x0)� f(x0)k "

2+ kfn0(x)� fn0(x0)k.

As the function fn0 is continuous, for any x0 in X, there exists ↵ > 0 such that

8x 2 X, d(x, x0) < ↵ =) kfn0(x)� fn0(x0)k <"

2·

We deduce that8x 2 X, d(x, x0) < ↵ =) kf(x)� f(x0)k < ",

which ends the proof of the proposition. ⇤

30

Remark. When X = N and E = K, which is R or C, we denote B(N,K) as `1(N,K). We mayomit K in this notation when it does not matter whether the sequences have real or complexvalues.

Exercise 2.1.1. We denote Cu(X,E) the set of uniformly continuous functions f in B(X,E).Prove that (Cu(X,E), k · kB(X,E)) is a Banach space.

Theorem 2.1.1. For p in the interval [1,1[, we consider `p(N), the set of all sequences x

taking values in K such that X

n2N|x(n)|p < 1.

We then set

kxk`pdef=⇣X

n2N|x(n)|p

⌘ 1p.

The space (`p(N), k · k`p) is a Banach space. Moreover, if p q, then `p(N) is a subspace

of `q(N), and the embedding ⇢

`p(N) �! `

q(N)x 7�! x

is a continuous linear map.

Proof. First, we prove that `p(N) is a vector space. Let x and y be two elements of `p(N).According to proposition 2.1.2, for any integer N , we have

NX

j=0

|x(j) + y(j)|p ✓⇣ NX

j=0

|x(j)|p⌘ 1

p+⇣ NX

j=0

|y(j)|p⌘ 1

p

◆p

�kxk`p + kyk`p

�p.

As a result, the series with general term |x(j) + y(j)|p converges, andX

j2N|x(j) + y(j)|p

�kxk`p + kyk`p

�p,

which ensures that (`p(N), k · k`p) is a normed space. Now we prove that it is a Banach space.Let (xq)q2N be a Cauchy sequence in `p(N). By definition,

8" > 0 , 9q" / 8q0 � q" , 8q0 � q" , kxq � xq0k`p(N) < " . (2.6)

As, for any p, |x(n)| kxk`p , then, for any n, the sequence (xq(n))q2N is a Cauchy sequencein K, and thus it converges to an element x(n) in K. It remains to prove that x = (x(n))n2Nis in `p(N), and that the sequence (xq)q2N converges to x in `p(N).

Using (2.6) with " = 1, we get

8q � q1 ,pkxqk`p 1 + kxq1k`p .

By definition of the `p norm, this implies that

8q � q1 ,p8N 2 N ,

NX

n=0

|xq(n)|p �1 + kxq1k`p

�p.

31

Taking the limit as q tends to infinity, we get

8N 2 N ,

NX

n=0

|x(n)|p �1 + kxq1k`p

�p,

which ensures that x is in `p(N). To prove convergence, we note that (2.6) implies that

8" > 0 , 9q" / 8q � q" , 8q0 � q" , 8N 2 N ,

NX

n=0

|xq(n)� xq0(n)|p < "p.

Taking the limit as q0 tends to infinity, we deduce that

8" > 0 , 9q" / 8q � q" , 8q0 � q" , 8N 2 N ,

NX

n=0

|xq(n)� x(n)|p "p.

Now, we make N tend to infinity, and we get

8" > 0 , 9q" / 8q � q" , 8q0 � q" ,

X

n2N|xq(n)� x(n)|p "

p.

Convergence is proved.To conclude the theorem, observe that, by equation (2.1), for any x in `p1(N), any p2 � p1

and any positive integer N ,

NX

n=0

⇣ |x(n)|kxk`p1

⌘p2

NX

n=0

⇣ |x(n)|kxk`p1

⌘p1

1.

We deduce that x belongs to `p2(N), and, by taking the limit in N , that

kxk`p2 kxk`p1 .

Using this on x � y, we see that the embedding is 1-Lipschitz continuous. The theorem isproved. ⇤

Exercise 2.1.2. Prove that the set of sequences that have only a finite number of non-zero

terms is dense in `p(N), for any p in the interval [1,1[.

Exercise 2.1.3. Show that, for any p in ]1,1[, the map x 7�! kxkp`p

is differentiable at any

point x in `p(N), and that, for any p in [2,1[, the map is twice differentiable.

2.2 Spaces of continuous linear maps

As we have endowed a normed space with a metric, we can define continuous functions onthem. Being vector spaces, linear maps obviously play an important role. The continuity ofsuch maps can be checked in a very simple way.

Theorem 2.2.1. Let E and F be two normed vector spaces, and ` be a linear map from E

to F . The following three statements are equivalent:

• the map ` is Lipschitz continuous,

32

• the map ` is continuous,

• there exists an open subset U of E such that ` is bounded on U , that is

supx2U

k`(x)kF < 1.

Proof. It is clear that the only point we need to prove is that the third condition implies thefirst one. Let U be an open set on which the linear map ` is bounded, and let x0 be a pointin U . The set U � x0 is an open set (exercise: prove it!) containing the origin. As it is open,there exists an open ball with centre 0 and radius ↵ which is included in U � x0. So, we canwrite that

Mdef= sup

x2B(0,↵)k`(x)k sup

x2U�x0

k`(x)k

supy2U

k`(y � x0)k

supy2U

k`(y)k + k`(x0)k.

Thus, the map ` is bounded on a ball with centre 0 and radius ↵. For any y in E \ {0}, wehave

↵

2kykEy 2 B(0,↵).

We deduce that ��`✓

↵

2kykEy

◆�� M.

Using the linearity of ` and the homogeneity of the norm, we see that

8y 2 E , k`(y)kF 2M

↵kykE .

This inequality also holds for y = x� x0, so the theorem is proved. ⇤

Proposition 2.2.1. Let E and F be two normed vector spaces; we denote L(E,F ) the set of

continuous linear maps from E to F . The mapping defined by

kLkL(E,F )def= sup

kxkE1kL(x)kF

is a norm on L(E,F ). If (F, k · kF ) is a Banach space, then (L(E,F ), k · kL(E,F )) is one too.

Proof. It is very easy to see that L(E,F ) is a vector space. Let us now check the threeproperties that define a norm. Assume that kLkL(E,F ) = 0. This implies that L(x) is equalto 0 for any x in the unit ball (which is what we usually call the ball with centre 0 andradius 1). Then, for any x 6= 0,

L

✓x

2kxkE

◆=

1

2kxkEL(x) = 0 .

We deduce that L(x) = 0, and therefore L = 0. Now, let L1 and L2 be two elements of L(E,F ).We have

kL1(x) + L2(x)kF kL1(x)kF + kL2(x)kF .

33

Thus, for any x in E with a norm less than 1, we have

kL1(x) + L2(x)kF kL1kL(E,F ) + kL2kL(E,F ).

The supremum being the smallest upper bound, we deduce that

kL1 + L2kL(E,F ) kL1kL(E,F ) + kL2kL(E,F ).

Finally, let L 2 L(E,F ) and � 2 K. We have

k�L(x)kF = |�| kL(x)kF ,

hence,sup

kxkE1k�L(x)kF = |�| sup

kxkE1kL(x)kF .

Thus, k · kL(E;F ) is a norm on L(E;F ).

Let us assume that (F, k · kF ) is complete, and let us consider a Cauchy sequence (Ln)n2Nin L(E,F ). By definition of the norm on L(E,F ), we deduce that, for any x in E, thesequence (Ln(x))n2N is a Cauchy sequence in F . As the space F is assumed to be complete,for every x in E, there exists an element in F , which we denote L(x) such that

limn!1

Ln(x) = L(x).

Now it suffices to prove that L is an element of L(E,F ), and that

limn!1

Ln = L in L(E,F ).

The uniqueness property for limits allows us to easily see that L is a linear map. As thesequence (Ln)n2N is a Cauchy sequence, it is bounded. So there exists a constant C such that

supn2N

kxkE1

kLn(x)kF C.

Taking the limit, we deduce that

supkxkE1

kL(x)kF C.

So the linear map L is continuous. We now prove the convergence of the sequence (Ln)n2Nto L in L(E,F ). As the sequence (Ln)n2N is a Cauchy sequence, for any positive ", thereexists an integer n0 such that, for any integer n � n0, we have

supkxkE1p2N

kLn(x)� Ln+p(x)kF < ".

Taking the limit as p tends to infinity, we get

supkxkE1

kLn(x)� L(x)kF ",


34

Proposition 2.2.2. Let E, F and G be three normed vector spaces, and (L1, L2) be an

element of L(E,F )⇥ L(F,G). Then the composition L2 � L1 belongs to L(E,G), and

kL2 � L1kL(E,G) kL1kL(E,F )kL2kL(F,G).

Proof. As the supremum is an upper bound, we have, for any x in the unit ball E,

kL2 � L1(x)kG kL2kL(F,G)kL1(x)kF kL2kL(F,G)kL1kL(E,F ).

Noting that the supremum is the smallest upper bound yields the result. ⇤

Remark. Even if F = K (the case of linear functionals), the supremum on the unit ball maynot be attained, as the following exercise, which we highly recommend, shows.

Exercise 2.2.1. Let (an)n2N be a sequence of elements in the interval ]0, 1[, which we assume

converges to 0. We define the linear form à on `1(N) by

hà, xidef=X

n2N(1� an)x(n).

Prove that kàk(`1(N))0 = 1, and that, for every x in the unit ball of `1(N), |hà, xi| is strictly

less than 1.

When E = F , the space L(E,F ) is denoted L(E). We are going to study this set, and, inparticular, its invertible elements. One of the basic results on invertible elements of L(E) isthe following.

Theorem 2.2.2. Let (E, k · k) be a Banach space. The set of elements in L(E) which have a

distance to Id strictly less than 1 are invertible in L(E). In other words,

8A 2 BL(E)(Id, 1) , 9!A�1 2 L(E) / A �A�1 = A�1 �A = Id .

Proof. We will use the following property.

Proposition 2.2.3. Let E be a Banach space, and (xn)n2N be a sequence of elements of E

such that X

n2NkxnkE < 1.

Then the sequence SN

def=

NX

n=0

xn converges.

Proof. The sequence (SN )N2N is a Cauchy sequence, due to

kSN+P � SNk N+PX

n=N+1

kxnk.

This proves the proposition. ⇤

35

Back to the proof of theorem 2.2.2. An element A in BL(E)(Id, 1) can be written as A = Id�L,with kLkL(E) strictly less than 1. Set

SN =NX

n=0

Ln.

Proposition 2.2.3 states that the sequence (SN )N2N converges to an element A�1 in L(E) as

soon as kLkL(E) is strictly less than 1. Also, we have

SNA = SNA = Id�LN+1

.

By taking the limit in the above equality, we conclude the proof of the theorem. ⇤

Corollary 2.2.1. The set U(E) of invertible elements in L(E) is open, and the mapping Inv,

defined by ⇢U(E) �! U(E)L 7�! L

�1

is continuous.

Proof. If L0 is a fixed element in U(E), we can write that

L = L0 � (L0 � L) =�Id�(L0 � L)L�1

0

�L0.

If we assume thatkL� L0kL(E) <

1

kL�10 kL(E)

,

we deduce from theorem 2.2.2 that Id�(L0 �L)L�10 is invertible, and therefore L is too. The

set of invertible elements is therefore open, because if L0 belongs to U(E), then the ball BL0

with centre L0 and radius kL�10 k�1

L(E) is included in U(E). Theorem 2.2.2 then implies that

L�1 = L

�10

1X

n=0

�(L0 � L)L�1

0

�n.

We deduce that, if L belongs to B(L0, ⇢0), with ⇢0 strictly less than kL�10 k�1

L(E), we have

��L�1 � L�10 kL(E) kL�1

0 kL(E)

1X

n=1

kL� L0knL(E)kL�10 kn

L(E)

kL� L0kL(E)kL�10 k2

L(E)

1X

n=0

kL� L0knL(E)kL�10 kn

L(E)

kL� L0kL(E)kL�10 k2

L(E)

1

1� ⇢kL�10 kL(E)

·

This ends the proof of the corollary. ⇤Remark. In fact, the proof shows that, when ⇢0 < kL�1

0 k�1L(E), the map L 7! L

�1 is k-Lipschitzcontinuous on BL(E)(L0, ⇢0), with

k =kL�1

0 k2L(E)

1� ⇢0kL�10 kL(E)

·

36

Exercise 2.2.2. Prove that the map Inv is in the class C1(U(E)).1

To conclude this section, we will characterise continuous multilinear maps. The followingtheorem is to multilinear maps what theorem 2.2.1 is to linear maps, a condition for continuity.

Theorem 2.2.3. Let ((Ej , k · kj))1jN be a family of normed vector spaces, and (F, k · k) be

a normed vector space. We consider an N -linear map from E1 ⇥ · · ·⇥ EN to F . This map f

is continuous from E = E1 ⇥ · · ·⇥ EN endowed with the norm

k(x1, · · ·xN )kE =X

1jN

kxjkEj

to F if and only if

supk(x1,··· ,xN )kE1

kf(x1, · · · , xN )kF < 1. (2.7)

Moreover, if f is continuous from E1 ⇥ · · · ⇥ EN to F , then it is Lipschitz continuous on all

bounded subsets of E1 ⇥ · · ·⇥ EN .

Proof. If f is continuous at the point 0, then there exists a positive real number ↵ such that

NX

j=1

kxjkEj < ↵ =) kf(x1, · · · , xN )kF < 1.

We then use a homogeneity argument. Let x = (x1, · · · , xN ) be a non-zero element of E witha norm in E less than 1. The norm of

↵

2x is therefore less than ↵. Thus,

kf(x1, · · · , xN )kF <

⇣ 2↵

⌘N

.

Conversely, assume that inequality (2.7) holds. Set h = (h1, · · · , hN ), and we denote h0def= 0,

and, for each j in {1, · · · , N}, ehjdef= (h1, · · · , hj , 0 · · · , 0). Then, for any fixed x in E, we have

f(x+ h)� f(x) =NX

j=1

f(x+ hj)� f(x+ hj�1).

Inequality (2.7) implies that, for any integer j between 1 and N , we have

kf(x+ hj)� f(x+ hj�1)kF C

⇣j�1Y

k=1

kxk + hkkEk

⌘khjkEj

⇣ NY

k=j+1

kxkkEk

⌘.

We can assume that khkE 1. Thus, for any integer j between 1 and N , we have

kf(x+ hj)� f(x+ hj�1)kF CxkhjkEj .

Taking the sum for j, we deduce that

kf(x+ h)� f(x)kF CxkhkE ,

and the theorem is proved. ⇤1This exercise demands the understanding of differential calculus in infinite dimensionnal spaces

37

2.3 Banach spaces, compactness and finite dimension

Whether the dimension of a space is finite or not has large implications on topology, as shownby the following statement.

Theorem 2.3.1. Let E be a normed vector space. If the dimension of E is finite, then the

closed unit ball is compact, and all norms on E are equivalent. Conversely, if the closed unit

ball of E is compact, then the dimension of the vector space E is finite.

Proof. Let E be a normed vector space with finite dimension N . We will prove that there is acontinuous linear bijection from RN endowed with the norm k · k1 to E, such that the inverseis also continuous. Consider (�!e j)1jN a basis of the space E, and the linear bijection I

defined by

I

8><

>:

RN �! E

x = (xj)1jN 7�!NX

j=1

xj�!e j .

The map x 7! I(x) is a linear bijection. Let us prove that it is continuous. We have

kI(x)kE NX

j=1

|xj | k�!e jkE

Mkxk1 with Mdef=

NX

j=1

k�!e jkE .

As the map I is linear, we deduce that

kI(x)� I(y)kE Mkx� yk1.

The functionx 7! kI(x)kE

is therefore continuous from RN to R+. By corollary 1.3.1 on page 23, the sphere SN�1, thatis the set of points x in RN such that kxk1 = 1, is compact. Moreover, as I is bijective, itdoes not vanish on SN�1. By corollary 1.3.2 on page 24, there exists a positive real number msuch that

8x 2 RN \ {0} , m ��I⇣

x

kxk1

⌘��E

M.

As the map I is linear, we have

8x 2 RN, mkxk1 kI(x)kE Mkxk1. (2.8)

So I and I�1 are continuous linear bijections. This implies that all norms on E are equivalent.

Therefore the open sets (resp. closed sets, compact sets) of E are precisely the images of open(resp. closed, compact) sets of RN through I. Thus, the closed unit ball of E is compact,as it is a closed subset of the image through I of the closed ball of RN with centre 0 andradius m

�1.The converse is based on the following geometric lemma.

38

Lemma 2.3.1. Let (E, k · k) be a normed vector space. We assume that there exists a real

number d in the interval ]0, 1[ and a finite family (xj)1jN such that

SEdef= {x 2 E , kxk = 1} ⇢

N[

j=1

B(xj , �).

Then the family (xj)1jN generates the space E.

If we accept the lemma for a moment, we only need to use the characterisation of com-pactness provided by theorem 1.3.4 on page 24 to conclude the proof. ⇤

Proof of lemma 2.3.1 Let F be the subspace of E generated by the family (xj)1jN . It is avector space with finite dimension, so, by the first part of our theorem, the space (F, k · k) iscomplete. It is therefore a closed subspace of E.

We now consider a vector y in E, and we will prove that it is a cluster point of F , whichis enough to prove the lemma. The key point we need to establish is the following:

8y 2 E , 9y1 2 F / ky � y1k �kyk. (2.9)

If y = 0, obviously y1 = 0. Let us assume that y is non-zero. This means that there exists j

in {1, · · · , N} such that ��y

kyk � xj

��

which we can rewrite ky � kykxjk �kyk, and this proves statement (2.9).We iterate this process. Assumption (2.9) implies that

9y2 2 F / ky � y1 � y2k �ky � y1k �2kyk.

By induction, we construct a sequence (yn)n2N of elements of F such that

8n 2 N ,

��y �nX

j=1

yj

�� nkyk.

It is clear that

limn!1

nX

j=1

yj = y

which proves that y is in F , because F is closed in E. The lemma is therefore proved. ⇤

Remark. In a vector space with finite dimension, and only in these, the compact subsetsare exactly the closed and bounded subsets. As the above theorem states, this is always falsein infinite-dimension spaces. To illustrate, let us consider the space E of continuous functionsfrom [0, 1] to R, endowed with the norm inducing uniform convergence. For every n 2 N, thefunction defined by fn(x)

def= x

n has norm 1, but the sequence (fn)n2N has no cluster point in

the space E, because

8x 2 [0, 1[ , limn!1

xn = 0 and lim

n!11n = 1.

39

2.4 Compactness in the space of continuous functions: Ascoli’stheorem

In finite dimension, and only in this setting, the compact subsets of a vector space areexactly the closed and bounded subsets. Theorem 2.3.1 below states that this is always falsein normed vector spaces with infinite dimension. To illustrate this, let us consider, in thespace E of continuous functions from [0, 1] to R, endowed with the norm inducing uniformconvergence, the sequence (fn)n2N defined by fn(x)

def= x

n. The norm of each element of thissequence is equal to 1, and yet, the sequence does not have a cluster point in the space E

because8x 2 [0, 1[ , lim

n!1xn = 0 and lim

n!11n = 1.

The aim of this section is to establish a criterion for compactness in the space of continuousfunctions from a compact metric space to a Banach space.

Theorem 2.4.1 (Ascoli). Let (X, d) be a compact metric space, and (E, k · kE) be a Banach

space. We consider a subset A of C(X,E) which satisfies the following two properties:

i) the subset A is uniformly equicontinuous, i.e.

8" > 0 , 9↵ > 0 / 8(x, x0) 2 X2, d(x, x0) < ↵ =) 8f 2 A , kf(x)� f(x0)kE < " ; (2.10)

ii) for any x 2 X, the closure of the set {f(x) , f 2 A} is compact.

Then, the closure of A is compact.

Proof. To prove this theorem, we will use the criterion stated in theorem 1.3.1 on page 20.Let " be a positive number. We will prove that we can cover A with a finite number of openballs with radius ", which, according to theorem 1.3.1 on page 20, yields the result.

By assumption (2.10), there exists a positive real number ↵ such that

8(x, x0) 2 X2, d(x, x0) < ↵ =) 8f 2 A , kf(x)� f(x0)kE <

"

3· (2.11)

The compactness of X ensures the existence of a finite sequence (xj)1jN of elementsof X such that

X =N[

j=1

B(xj ,↵).

Set Aj

def= {f(xj) , f 2 A}. By assumption, the closure of Aj , Aj , is compact in E. By

proposition 1.3.4 on page 21, the product A1 ⇥ · · ·⇥AN is compact in the Banach space EN ,

endowed with the normk(yj)1jNk1

def= max

1jN

kyjkE .

So the closure of the product A1 ⇥ · · · ⇥ AN is compact in the Banach space EN . As the

subset AN of EN defined by

A def=�(f(x1), · · · , f(xN )) , f 2 A

is included in the product A1 ⇥ · · · ⇥ AN , so its closure is also compact. By theorem 1.3.1on page 20, there exists a finite sequence (fk)1kM of elements of A such that the balls (for

40

the norm k · k1 we defined above) with centre (fk(xj))1jN and radius "/3 cover A, whichmeans that

8f 2 A , 9k 2 {1, · · · ,M} / 8j 2 {1, · · · , N} , kf(xj)� fk(xj)kE <"

3·

We now prove that the balls in C(X,E) with centres fk and radius " cover A. To do so,let us consider f , an element of A. There exists an index k in {1, · · · ,M} such that

8j 2 {1, · · · , N} , kf(xj)� fk(xj)kE <"

3·

Let x be a point in X. There exists an index J in {1, · · · , N} such that d(x, xj) < ↵. We cantherefore write

kf(x)� fk(x)kE kf(x)� f(xj)kE + kf(xj)� fk(xj)kE + kfk(xj)� fk(x)kE< " .

This proves the theorem. ⇤Exercise 2.4.1. Prove the converse of Ascoli’s theorem: if A is a compact subset of C(X,E),then it satisfies conditions i) and ii) of Ascoli’s theorem. The equivalence is known as the

Arzelà-Ascoli theorem.

Exercise 2.4.2. Let (X, d) be a compact metric space, and Y be a compact subset of a Banach

space (E, k · kE). Prove that the set of Lipschitz continuous functions with a given Lipschitz

constant k from X to Y is compact in C(X,E).

Exercise 2.4.3. Give an example of a sequence of functions (fn)n2N from [0, 1] to R which

is bounded in the space of Lipschitz continuous functions, and which converges to 0 in the

space C([0, 1],R) but not in the space of Lipschitz continuous functions from [0, 1] to R.

2.5 About the Stone-Weierstrass theorem

The aim of this section is to find dense subspaces of the space of continuous functions on acompact metric space (X, d). We begin with the case where X is the interval [0, 1].

Theorem 2.5.1 (Bernstein). Let f be a continuous function from the interval [0, 1] to a

Banach space E. The sequence of functions (Sn(f))n2N defined by

Sn(f)(x)def=

nX

k=0

f

⇣k

n

⌘C

k

nxk(1� x)n�k

,

in which the Ckn are the binomial coefficients, is uniformly convergent to f on the interval [0, 1].

Proof. For any x in the interval [0, 1] and any positive integer n, we have 1 = (x+ 1� x)n, sothe binomial formula implies that

nX

k=0

Ck

nxk(1� x)n�k = 1 (2.12)

By definition of the sequence Sn(f), we deduce that

f(x)� Sn(f)(x) =nX

k=0

⇣f(x)� f

⇣k

n

⌘⌘C

k

nxk(1� x)n�k

.

41

By Heine’s theorem, theorem 1.3.3 on page 23, the function f is uniformly continuous on thecompact interval [0, 1], so, if we consider an arbitrary positive number ", there exists a positivereal number ↵" such that

|x� y| < ↵" =) kf(x)� f(y)kE <"

2·

We can then write

kf(x)� Sn(f)(x)kE nX

k=0

��f(x)� f

⇣k

n

⌘��E

xk(1� x)n�k

"

2+ 2 sup

x2[0,1]kf(x)kE

X

|k�nx|�n↵"

Ck

nxk(1� x)n�k

.

By multiplying and dividing by (k � nx)2, we get

kf(x)� Sn(f)(x)kE "

2+ sup

x2[0,1]kf(x)kE

X

|k�nx|�n↵

(k � nx)2

(k � nx)2C

k

nxk(1� x)n�k

"

2+

2

n2↵2sup

x2[0,1]kf(x)kE

X

|k�nx|�n↵"

(k � nx)2Ck

nxk(1� x)n�k

"

2+

2

n2↵2sup

x2[0,1]kf(x)kE

nX

k=0

(k � nx)2Ck

nxk(1� x)n�k

. (2.13)

We can compute the sum on the right. By differentiating equality (2.12), then multiplyingby x, we get

nX

k=0

kCk

nxk(1� x)n�k � nx

n�1X

k=0

Ck

n�1xk(1� x)n�1�k = 0,

which, by using (2.12) again, yields thatnX

k=0

kCk

nxk(1� x)n�k = nx. (2.14)

Then, by differentiating and multiplying by x the above equality, we get

nX

k=0

k2C

k

nxk(1� x)n�k � nx

n�1X

k=0

kCk

n�1xk(1� x)n�1�k = nx

Using equality (2.14) in this leads tonX

k=0

k2C

k

nxk(1� x)n�k = nx(1 + (n� 1)x)

which implies thatnX

k=0

(k � nx)2Ck

nxk(1� x)n�k =

nX

k=0

k2C

k

nxk(1� x)n�k � 2nx

nX

k=0

kCk

nxk(1� x)n�k + n

2x2

= nx(1 + (n� 1)x)� 2n2x2 + n

2x2

= nx(1� x).

42

Equality (2.13) then becomes

kf(x)� Sn(f)(x)kE "

2+

2

n↵2sup

x2[0,1]kf(x)kE .

All that remains to do is to choose n such that n � n"def=h

supx2[0,1]

kf(x)kE4

"↵2"

i+ 1, and the

theorem is proved. ⇤We are going to state a criterion for density for sub-algebras of C(X,K), where (X, d) is a

compact metric space. First, here is a definition.

Definition 2.5.1. Let (X, d) be a metric space and A a subset of C(X,K). We say that the

subset A separates the points of X if and only if for any pair (x, y) in X2

such that x 6= y,

there exists a function f in A such that f(x) 6= f(y).

We can now state the following theorem on density.

Theorem 2.5.2 (Stone-Weierstrass). Let (X, d) be a compact metric space, and A be a sub-

algebra of the vector space C(X,R). If A contains all the constant functions from X to R, and

separates the points of X, then A is a dense subset of C(X,R).

The complex-valued case differs slightly.

Theorem 2.5.3. Let (X, d) be a compact metric space, and A be a sub-algebra of the vector

space C(X,C). If A contains all the constant functions from X to C, separates the points

of X, and is stable by conjugation (i.e. f 2 A ) f 2 A), then A is a dense subset of C(X,C).

We can deduce the following result.

Corollary 2.5.1. If X is a compact space, then the Banach space C(X,K) is separable, which

means that there exists a dense sequence of points of C(X,K).

The proof of the general Stone-Weierstrass theorem is given for the reader’s culture.Proof of theorem 2.5.2.

One of the main steps of the proof is the following lemma, which we will accept without proof forthe moment.

Lemma 2.5.1. Let A be a sub-algebra of C(X,R). For any function f in A, the function |f | belongsto A.

Remark. Since max{f, g} =1

2(f + g+ |f � g|) and min{f, g} =

1

2(f + g� |f � g|), if f and g are two

functions that belong to A, then their minimum and maximum are in A.

Back to the proof of theorem 2.5.2. The fact that A separates the points of X implies that

(S+) 8(x1, x2) 2 X2/px1 6= x2 , 8(↵1,↵2) 2 R2

, 9h 2 A/ h(xj) = ↵j .

For the same reason, there exists g in A such that g(x1) 6= g(x2). We then set

h(z)def= ↵1 + (↵2 � ↵1)

g(z)� g(x1)

g(x2)� g(x1)·

Let us consider f a function in C(X,R), x0 a point of X and " a positive real number. We can provethat

9hx0 2 A/ hx0(x0) = f(x0) and 8z 2 X , hx0(z) f(z) + ". (2.15)

43

Indeed, by statement (S+), for any y in X, there exists a function hx0 in A such that h coincideswith f at the points x0 and y. Both functions f and hx0,y are continuous, so there exists a positivenumber ↵x0,y such that

d(z, y) < ↵x0,y =) hx0,y(z) < f(y) + ".

As the space (X, d) is assumed to be compact, it can be covered with a finite number of balls like theone described above, which means that there exists a finite sequence (yj)1jN such that

X =N[

j=1

B(yj ,↵x0,yj ).

Set hx0

def= min

1jN

hx0,yj . By lemma 2.5.1, the function hx0 belongs to A and satisfies, for every z

in X, hx0(z) < f(z) + ". Thus, (2.15) is proved.Therefore, for any x 2 X, there exists a function hx 2 A such that

hx(x) = f(x) and 8y 2 X , hx(y) < f(y) + ".

The functions hx and f are continuous, so, for any y 2 X, the exists a positive number ↵x such that

8z 2 B(x,↵x) , hx(z) > f(z)� ".

Again, X can be covered by a finite number of such balls, so there is a finite family (xj)1jN of pointsof X such that

X =N[

j=1

B(xj ,↵xj ).

Let us set hdef= max

1jN

hxj . By lemma 2.5.1, h belongs to A, and, moreover, it satisfies

8y 2 X , f(y)� " < h(y) < f(y) + ",

which means that kf�hkC(X,R) < ". The theorem is therefore proved, providing we prove lemma 2.5.1.⇤

Proof of lemma 2.5.1. Let us first note that we can assume that A is closed, because if A is a sub-algebra of C(X,K), then so is A (exercise: prove it!). The proof then relies on the following lemma,which we accept for the time being.

Lemma 2.5.2. There exists a sequence of polynomial functions (Pn)n2N such that

limn!1

Pn =p· in the space C([0, 1];R).

Let f be an element of A, and let us set

fndef= Pn

⇣f2

kfk

⌘where, of course, kfk = sup

x2X

|f(x)|.

The sequence (fn)n2N is a sequence of elements of A because A is an algebra. By lemma 2.5.2, thesequence of functions (fn)n2N converges to kfk�1

pf2 = kfk�1|f |. Thus, |f | is in A. This means that

theorem 2.5.2 is now proved providing we show lemma 2.5.2. ⇤

Proof of lemma 2.5.2. We define by induction the sequence of polynomial functions

Pn+1(t) = Pn(t) +1

2(t� P

2n(t)) and P0 = 0.

44

We will show by induction that, for every t 2 [0, 1], 0 Pn(t) pt. This is true for P0. Let us assume

that it is true for Pn. The polynomial function Pn+1 is non-negative on [0, 1] as it is the sum of twonon-negative functions. Moreover, as Pn is non-negative on [0, 1],

t� P2n+1(t) = t�

⇣Pn(t) +

1

2(t� Pn(t))

2⌘2

� t� P2n(t)

� 0

As Pn+1(t)�Pn(t) = 1/2(t�P2n(t)) for any t 2 [0, 1], we have Pn+1(t) � Pn(t). Thus, for any t 2 [0, 1],

the sequence (Pn(t))n2N is non-decreasing and bounded from above bypt, so it converges. Let `(t)

denote the limit. Taking the limit in the relation between Pn and Pn+1, we see that t� `2(t) = 0, and

the fact that this convergence is in fact uniform is given by the following theorem.

Theorem 2.5.4 (Dini). Let (fn)n2N be a sequence of elements of C(X,R) such that, for any x in X,the sequence (fn(x))n2N is a non-decreasing sequence with an upper bound. If g(x) def

= limn!1

fn(x) isa continuous function on X, then the sequence (fn)n2N converges uniformly to g, that is

limn!1

supx2X

|fn(x)� g(x)| = 0.

Proof. For any x 2 X, the sequence (fn(x))n2N is non-decreasing and bounded from above, so, forany " > 0, there exists an integer nx such that

g(x)� "

2< fnx(x).

As both functions g and fnx are continuous on the compact space (X, d), they are uniformly continuouson (X, d). Thus, there exists a positive real number ↵x such that

d(y, y0) < ↵x =) |g(y)� g(y0)|+ |fnx(y)� fnx(y0)| < "

2·

The family (B(x,↵x))x2X covers the compact space (X, d), so we can extract a finite sub-family thatalso covers X, which means that there exists a sequence (xj)1jN such that

X =N[

j=1

B(xj ,↵xj ).

Set n0def= max

1jN

nxj . The sequence (fn(y))n2N is non-decreasing, so, for any j in {1, · · · , N}, we have

fn0(y) � fnxj(y)

� fnxj(xj)� |fnxj

(y)� fnxj(xj)|

� g(xj)�"

2� |fnxj

(y)� fnxj(xj)|

� g(y)� "

2� |g(y)� g(xj)|� |fnxj

(y)� fnxj(xj)|.

Let j be such that y 2 B(xj ,↵xj ). We get

fn0(y) � g(y)� ".

As the sequence (fn(y))n2N is non-decreasing, we obtain

n � n0 =) 8y 2 X , g(y) � fn(y) > g(y)� ".

The theorem is proved. ⇤

45

2.6 Notions on separable metric spaces

The notion of separable spaces is based on countability.

Definition 2.6.1. Let X be a set. The set X is said to be countable if and only if there exists

an one to one map from X to N.

Obviously, any set with finite cardinality is countable. Moreover, we have the followingproposition.

Proposition 2.6.1. Let X be a countable set with infinite cardinality (also called countably

infinite). Then there exists a bijective map from X to N.

Proof. By assumption, there exists an one to one function f from X to N. Therefore it is abijective function from X to Y = f(X). It suffices then to show that an infinite part Y of Nis in one-to-one correspondance with N. We define by induction the sequence

b(n)def= minY \ {b(0), · · · , b(n� 1)}

The function b is increasing, and therefore one to one. It is bijective because the elements of Yare taken in ascending order. ⇤

Proposition 2.6.2. Any finite product of countable sets is countable.

Proof. It suffices to show the result for the product of two sets, which, in the same way as inthe previous proposition, means we must construct an one to one function from N ⇥ N to N.We define

�

8<

:N⇥ N =) N

(p, q) 7�! p+(p+ q + 1)(p+ q + 2)

2·

Let us prove that � is one to one. To do so, we start by observing that if (p, q) and (p0, q0) aretwo pairs of N2 such that p

0 + q0 is strictly greater than p+ q, then �(p0, q0) is strictly greater

than �(p, q). Indeed, we have p0 + q

0 greater than p+ q + 1, so

(p0 + q0 + 1)(p0 + q

0 + 2) � (p+ q + 3)(p+ q + 2)

� 2(p+ q + 2) + (p+ q + 1)(p+ q + 2).

We deduce that�(p0, q0) � p

0 + q + 2 + �(p, q).

Thus, if �(p, q) = �(p0, q0), then p + q = p0 + q

0, which implies immediately that p = p0

and q = q0, proving that � is one to one. ⇤

Corollary 2.6.1. The sets Z and Q are countable.

Corollary 2.6.2. A countable disjoint union of countable sets is countable. To be precise,

let An be a sequence of countable sets, and we set

A = {(n, a) with a 2 An}.

The set A is countable.

46

Proof. Each set An is countable. So there exists, for each n, an one to one map fn from An

to N. We then define

�

⇢A �! N2

a = (n, an) 7�! (n,�n(an)).

This map is one to one. Indeed, if �(a) = �(b), then this means that (n, an) = (m, bm),so n = m and an = bm. Now, let � be the map from proposition 2.6.1: the map � � � is oneto one from A to N as it is the composition of two one to one maps. ⇤

We have the following negative result.

Theorem 2.6.1. Let (X, d) be a complete metric space with infinite cardinality, such that

the interior of each singleton is empty. Then the set X is not countable.

Proof. Let (xn)n2N be a sequence of elements of X (since all countable set can be representedthis way). Baire’s theorem, as stated in theorem 1.2.3, yields that the interior of the set

[

n2N{xn}

is empty, so it cannot be equal to X. ⇤We can now define the notion of separable metric spaces.

Definition 2.6.2. Let (X, d) be a metric space. The space (X, d) is said to be separable if

and only if there exists a dense sequence of elements of X.

In other words, this definition means that, in a separable metric space (X, d), there existsa sequence (xn)n2N of elements of X such that

8x 2 X , 8" > 0 , 9m/ d(x, xm) < ".

Exemple. The space R endowed with the metric d(x, y) = |x � y| is separable, because thesequence of all rationals is dense in R.

Proposition 2.6.3. Let (X1, d1), · · · , (XN , dN ) be a family of N separable metric spaces. If

we set


dj(xj , yj),

then the metric space (X, d) is separable.

Proof. In each space Xj , there exists a countable subset Dj which is dense in Xj . Consider x =(x1, · · · , xN ) a point in X, and set D = D1 ⇥ · · ·⇥DN . As it is a finite product of countablesets, D is a countable subset of X, and, as Dj is dense in Xj , for every positive ", and forevery integer j between 1 and N , there exists yj in Dj such that

dj(xj , yj) < ".

By definition of the metric d, we have d(x, y) < ". The proposition is proved. ⇤

Proposition 2.6.4. Let (X, d) be a separable metric space, and A be a subset of X. Then

the metric space (A, d) is separable.

47

Proof. To prove this proposition, we denote D a countable dense subset of X. For any integer nand any x in D, we choose an element ax,n in B(x, n�1) \ A, if this set is not empty. If itis empty, we select any element of A. We have thus constructed a countable subset of A,eD def

= (axn)(x,n)2D⇥N.Let us prove that eD is dense in A. Let a be an element in A. There exists an element x

in D such that d(x, a) < n�1. By definition of eD, the element ax,n is in eD \ B(x, n�1), so we

have thatd(a, ax,n) d(a, x) + d(x, ax,n) <

2

n

,

hence the proposition. ⇤

Theorem 2.6.2. All compact metric spaces (X, d) are separable.

Proof. Let Bn be the set of balls with radius n�1. Of course, this set covers X with

open balls, and, given that X is compact, for every integer n, there exists a finite familyof points (xj,n)1jJ(n) such that

X =

J(n)[

j=1

B(xj,n, n�1).

Let D be the set defined by

D = {xj,n , 1 j J(n) , n 2 N}.

Corollary 2.6.2 implies that D is countable. Let us prove that it is dense. To do so, consideran arbitrary positive real number " and any point x in X. We then choose an integer n suchthat n is greater than the inverse of ". By definition of the sequence (xj,n), there exists j suchthat

d(x, xj,n) <1

n·

Given that n is arbitrary, the theorem is proved. ⇤

We end the chapter with an example of a non-separable metric space.

Proposition 2.6.5. The space `1(N) is not a separable.

Proof. It relies on the following lemma.

Lemma 2.6.1. Let (X, d) be a metric space. If there exists an uncountable subset A of X,

and a positive number ↵ such that

8(x1, x2) 2 A2, x1 6= x2 ) d(x1, x2) � ↵,

then the metric space (X, d) is not separable.

Proof. Let D be a dense subset of X. For any a in A, there exists z in D such that

d(a, z) ↵

3·

48

Let a1 and a2 be two different elements of A. They correspond to two elements z1 and z2 in D

which satisfy the above inequality. As a result, we have

↵ d(a1, a2)

d(a1, z1) + d(a2, z2) + d(z1, z2)

↵

3+↵

3+ d(z1, z2)

2↵

3+ d(z1, z2).

Hence z1 cannot be equal to z2. So a dense subset of X cannot be countable, which provesthe lemma. ⇤Back to the proof of Proposition 2.6.5 Let us observe that for any couple (A1, A2) of subsetsof N such that A1 6= A2. We have

k1A1 � 1A2k`1(N) = 1.

It is enough to observe that the set P(N) of the subsets of N is not countable. It is a consequenceof the following lemma, which concludes the proof of Proposition 2.6.5. ⇤

Lemma 2.6.2. Let X be a set. There is no onto map between X and the set P(X) the set ofthe subsets of X.

Proof. Let us consider a map f from X onto P(X). We shall prove by contradiction that f isnot onto. Let us consider the set

F =�x 2 X / x 62 f(x)

.

Let us assume that there exists an element x of X such that f(x) = F . If x belongs to F thenby definition of F , x does not belong to f(x) = F . Thus x does not belong to F . This leadsto contradiction. ⇤

49

Chapter 3

Duality in Banach spaces

Introduction

This short chapter is very important. It deals with a particular type of Banach space: thedual space (or topological dual space) of a Banach space, which is the set of continuous linearfunctionals on that space. First, we present the general concept of transposing a linear map.We point out that, in infinite dimension, this is far from simply transposing a matrix. Astriking new fact we will discover in this chapter is that the dual of a Banach space may bevery different from that space. This is of course not the case in finite dimension, where thedimension of the dual space is equal to that of the original vector space.

In the second section, we will give a precise mathematical meaning to the sentence “thedual space of E is F ”. This definition is fundamental. It allows one to construct isomorphismsbetween the dual E0 of a Banach space and another Banach space F , which gives us a de-scription of E0. The duals of the spaces of sequences which are summable when elevated to apower p will be studied in detail as an essential example.

The third section introduces a crucial concept, which is that of a weaker notion of conver-gence for a sequence of elements in the dual of a Banach space. It is the concept of weak-?convergence, which is essentially pointwise convergence. This allows one to extract subse-quences from any bounded sequence in the dual space which converge in this sense. Hence theimportance of the identification theorems proved beforehand: if a space can be identified asthe dual space of another, then it will inherit this property. We will see an application of thisresult in the next chapter, when we will prove theorem 4.4.2 on page 75, in which we studythe structure of self-adjoint compact operators.

3.1 A presentation of the concept of duality

The reader should already be well informed on the notion of dual vector spaces. Nonetheless,we briefly recall some fundamental points. If E is any vector space, we denote E

? the spaceof linear forms defined on E. Moreover, if ` is a linear form on E and x is a vector in E, wedenote

h`, xi def= `(x).

The first key notion is that of the transpose of a linear map. Consider a linear map L from E

to E. Its transpose is the map from E? (the set of linear functionals on E) to E

? defined by

htL(`),�!h i = h`, L

�!h i.

51

Let us note that if L1 and L2 are two linear maps from E to E, then

t(L1 � L2) =tL2 � t

L1.

Indeed, by definition of the transpose of a map, we have

ht(L1 � L2)(`),�!h i = h`, L1 � L2(

�!h )i

= h`, L1(L2(�!h ))i

= htL1(`), L2(�!h )i

= htL2(tL1(`)),

�!h i.

Hence the result.

In the case where the vector space E has finite dimension, we can choose (�!e j)1jN , abasis of E. We can easily prove that the family of linear forms (e?

j)1jN , defined by

he?j ,�!e ki = �j,k

forms a basis of the vector space of linear forms E0, and is called a dual basis. Moreover,

it is easy to see that if the linear map L is represented by the matrix (Li

j)1i,jN in a ba-

sis (�!e j)1jN of E, then its transpose tL is represented by the matrix (eLi

j)1i,jN , with

eLi

j= L

j

i, in the dual basis (e?

j)1jN . Thus we see that, if E has finite dimension, then the

set of continuous linear forms (note that linear forms in this case are always continuous) is avector space with the same dimension.

In this chapter, we will study the case where E is a general Banach space.

Definition 3.1.1. We call the topological dual space (dual for short) of a normed K vector

space (E, k · k), denoted E0, the vector space of continuous linear functionals on E. It is

endowed with the norm

k`kE0

def= sup

kxk1|h`, xi|.

Compared with the finite dimension case, a radically new phenomenon appears. As we willsee after the proof of theorem 3.2.1, the Banach space `1(N) of absolutely convergent series isseparable, while its dual is not. At this stage, this sounds like it will create some difficulties,but transposing in this setting will in fact prove to be extremely useful. This idea is at theheart of the notion of quasi-derivatives that we will introduce in chapter 6 and, more generally,in the theory of distributions, which we will study in chapter 8.

There exists a natural bilinear form defined on the space E0 ⇥ E.

Proposition 3.1.1. Let E be a normed vector space. Then the mapping h·, ·i defined by

h·, ·i⇢

E0 ⇥ E �! K(`, x) 7�! h`, xi.

is a continuous bilinear form.

Proof. By definition of the norm on E0, we have |h`, xi| k`kE0kxk. Theorem 2.2.3 on page 37,

characterising continuous multilinear maps, then ends the proof. ⇤

52

We are now going to present the transpose of a linear map. This notion will play a centralrole in chapter 8.

Proposition 3.1.2. Let E and F be two normed vector spaces, and let A be an element of

the space L(E,F ). The maptA, defined by

⇢F

0 �! E0

` 7�! tA(`) : x 7! h`, Axi

is a continuous linear map.

Proof. We can write

|h`, Axi| k`kF 0kAxkE k`kF 0kAkL(E,F )kxkE .

Hence, we have

ktA(`)kE0 = supkxkE1

|h`, Axi|

k`kF 0kAkL(E,F ).

The proposition is proved. ⇤

Definition 3.1.2. The linear map defined above is called the transpose of the linear map A.

Proposition 3.1.3. Let E, F and G be three normed vector spaces, and let (A,B) be an

element of L(E,F )⇥ L(F,G). Then,

t(B �A) = tA � t

B

Moreover, if E = F , then

A 2 U(E) () tA 2 U(E0) and (tA)�1 = t(A�1).

Proof. The first equality is purely algebraic. Indeed, for any linear form ` on G, and for any x

in E, we have

htA � tB(`), xi = htB(`), Axi

= h`, B �Axi= ht(B �A)(`), xi.

Then, if A is in U(E), we have A �A�1 = A�1 �A = IdE . By transposing this equality, we get

tA � t(A�1) = t(A�1) � t

A = IdE0 .

Hence, tA is an invertible element of L(E0), and (tA)�1 = t(A�1). The proposition is proved.

⇤

53

3.2 Identifying a normed space as a dual space

This section is very important for the remainder of the course, in particular the chapter ondistribution theory. It is common to say that a given space “is the dual” of another. We needto give a precise meaning to this sentence.

Proposition 3.2.1. Let E and F be two Banach spaces, and let B be a bilinear continuous

functional on F ⇥ E. The mapping �B defined by

�B

⇢F �! E

0

y 7�! �B(y) : h�B(y), xi = B(y, x)(3.1)

is in L(F,E0).

Proof. As the bilinear form B is continuous, we can write that

|h�B(y), xi| CkykF kxkE .

By definition of the norm on E0, we have k�B(y)kE0 CkykF ; hence the proposition. ⇤

Definition 3.2.1. Let E and F be two Banach spaces, and let B be a bilinear continuous

functional on F ⇥ E. We say that B identifies E0with F if and only if the map �B defined

by (3.1) is an isomorphism from F to E0.

In other words, this means that there exists a constant C such that, for any continuouslinear functional ` on E, there exists a unique element y in F such that

8x 2 E , h`, xi = B(x, y) and C�1kykF k`kE0 CkykF .

Theorem 3.2.1. Let p be a number in the interval [1,+1[, and set E = `p(N) and F = `

p0

(N),where p

0 = p/(p� 1). We consider the following bilinear map B, defined by

B

8<

:

`p0

(N)⇥ `p(N) �! K

(y, x) 7�! B(y, x)def=X

n2Nx(n)y(n).

The bilinear form B identifies (`p(N))0 with `p0

(N). Moreover, the map �B induced by propo-

sition 3.2.1 is an isometry which means

kyk`p

0 (N) = supkxk`p(N)1

��B(y, x)| .

Proof. Hölder’s inequality states that

|B(y, x)| kxk`p(N)kyk`p0 (N). (3.2)

So the bilinear form B is continuous. Let us prove that the map �B is a bijection. Consider Ta linear functional on `p(N), and let us look for an element y in `p0(N) such that

8x 2 `p(N) , hT, xi = B(y, x) = h�B(y), xi. (3.3)

If the above is satisfied by y, then it must be satisfied for x = en, where en denotes the sequencewith all terms equal to zero, except the one with index n, which is equal to 1. Therefore, ymust verify

y(n) = hT, eni. (3.4)

54

This proves that �B is one to one. Moreover, by linearity, if V denotes the set of finite linearcombinations of elements of the sequence (en)n2N, then, for any x in V ,

B(y, x) = hT, xi. (3.5)

We must now prove that the sequence y defined by (3.4) is an element of `p0(N).Let us first assume that p is equal to 1. For any integer n, we have

|y(n)| kTk(`1(N))0kenk`1(N) kTk(`1(N))0 .

So the sequence y = (y(n))n2N is in `1(N), and we effectively have kyk`1(N) kTk(`1(N))0 .

Both terms in Equality (3.5) are continuous and coincide on a dense subspace of `1(N). Thisends the proof of the theorem for p = 1.

Now, let us assume that p is strictly greater than 1. We consider the sequence (xN )N2N ofelements of `p(N) defined by

xNdef=X

nN

y(n)

|y(n)| |y(n)|1

p�1 en.

This implies that

xN (n) =y(n)

|y(n)| |y(n)|1

p�1 if n N, and 0 otherwise.

We therefore have X

n2N|xN (n)|p =

X

nN

|y(n)|p0 .

B(y, xN ) =NX

n=0

y(n)

|y(n)| |y(n)|1

p�1B(en, y)

=NX

n=0

|y(n)|1

p�1+1

=NX

n=0

|y(n)|p0 .

Moreover, we know that

B(y, xr) =NX

n=0

y(n)

|y(n)| |y(n)|1

p�1T (en) = hT, xN i.

Since the linear form T is continuous, we haveX

nN

|y(n)|p0 kTk(`p(N))0kxNk`p(N)

kTk(`p(N))0⇣X

nN

|y(n)|p0⌘ 1

p.

55

Hence, we get⇣X

nN

|y(n)|p0⌘1� 1

p kTk(`p(N))0 .

As this inequality is true for any integer r, the sequence y is indeed in `p0(N), and we have

kyk`p

0 (N) kTk(`p(N))0 . (3.6)

The subspace V is dense in `p(N), so we observe again that equality (3.5) states that two

continuous linear functionals coincide on a dense subspace. Hence, they are equal on theentire space, and this ends the proof of the theorem. ⇤

Exercise 3.2.1. Let c0(N) denote the Banach space defined by

c0(N)def= {x = (x(n))n2N / lim

n!1x(n) = 0} endowed with the norm kxk`1(N)

def= sup

n2N|x(n)|.

Prove that c0(N) is a closed subspace of `1(N). Then, by following the ideas of the proof we

have just completed, prove that the bilinear form B identifies c0(N)0 with `1(N).

Remark. At the start of this chapter, we noted that, in the case of a vector space E withfinite dimension, the dual space is a vector space with the same dimension, and thereforea space that is isomorphic to E. No such luck in infinite dimension: the previous theoremshowed that the dual space of `1(N) is isometric to `1(N), and thus the space (`1(N))0 hasthe same topological properties as `1(N). In particular, it is not separable. However, we canshow that `1(N) is separable. As an exercise, prove that the space of sequences which have allterms, apart from a finite number of them, equal to zero, is dense in `1(N) (and, in fact, densein `p(N) for any p in [1,+1[).

Exercise 3.2.2. Let s be a real number. We define the space `2,s(N) by

`2,s(N) def

=�u = (u(n))n2N / ((1 + n

2)s2u(n))n2N 2 `

2(N) ,

endowed with the norm

kuk`2,s(N)def=

X

n2N(1 + n

2)s|u(n)|2! 1

2

.

Let B1 and B2 be two bilinear functionals defined as follows:

B1

8<

:

`2,s(N)⇥ `

2,s(N) �! K(y, x) 7�! B1(y, x)

def=X

n2N(1 + n

2)sx(n)y(n).

B2

8<

:

`2,�s(N)⇥ `

2,s(N) �! K(y, x) 7�! B2(y, x)

def=X

n2Nx(n)y(n).

1) Prove that the bilinear form B1 identifies (`2,s(N))0 with `2,s(N), and that B2 identi-

fies (`2,s(N))0 with `2,�s(N).

2) Propose a onto linear isometry from `2,s(N) to `

2,�s(N).

56

3.3 A weaker sense for convergence in E0

Definition 3.3.1. Let E be a normed vector space. We consider a sequence (`n)n2N of

elements of E0, and ` an element of E

0. We say that the sequence (`n)n2N is weak-? convergent

to `, denoted limf?n!1

`n = `, if and only if

8x 2 E , limn!1

h`n, xi = h`, xi.

First, we must note that weak-? convergence is implied by convergence in norm (the usualsense). Indeed, for any x in E, we have

��h`n, xi � h`, xi�� =

��h`n � `, xi��

k`n � `kE0kxkE .

Let us now show a sequence (`n)n2N of E0 which is weak-? convergent, but not convergentin norm. Consider the space `1(N) and the sequence (`n)n2N of linear functionals defined by

h`n, xidef= x(n).

For any x in `1(N), lim

n!1x(n) = 0. Hence, the sequence (`n)n2N is weak-? convergent to 0.

However, k`nk(`1(N))0 = 1, so the sequence does not converge to 0 in norm.

We will now show some properties of weak-? convergence.

Theorem 3.3.1. Let E be a Banach space, and (`n)n2N be a sequence of elements of E0.

Assume that limf?n!1

`n = `. Then the sequence (`n)n2N is bounded in E0, which implies that `

is an element of E0. Moreover, we have

k`kL(E,K) lim infn!1

k`nkL(E,K).

Proof. This is merely a statement of the classic Banach-Steinhaus theorem in the setting oflinear functionals. We will prove the general theorem, which is the following.

Theorem 3.3.2 (Banach-Steinhaus). Let E and F be two normed spaces, and let (Ln)n2Nbe a sequence in L(E,F ). Assume that E is complete, and that, for any x in E, the se-

quence (Ln(x))n2N has a limit, denoted L(x).Then the sequence (Ln)n2N is bounded in L(E,F ), so the map L is also in L(E,F ). Also,

we have

kLkL(E,F ) lim infn!1

kLnkL(E,F ).

Proof. The key ingredient is the following lemma.

Lemma 3.3.1. Let E and F be two normed spaces, and let (Ln)n2N be a sequence in L(E,F ).Assume that E is complete, and that, for any x in E, the sequence (Ln(x))n2N is bounded

in F .

Then, the sequence (Ln)n2N is bounded in L(E,F ).

Proof. Let us consider the sets Fn,p defined by

Fn,p = {x 2 E / kLn(x)kF p}.

57

These sets are closed, as they are preimages of closed sets by a continuous map. So, the sets Fp

defined byFp

def=\

n2NFn,p

are also closed, because they are intersections of closed sets. Moreover, let x be an elementof E. As the sequence (Ln(x))n2N is assumed to be bounded, there exists an integer p suchthat x belongs to Fp. This means that the union of the closed sets Fp is equal to the entirespace E. By corollary 1.2.1 on page 17, deduced from Baire’s theorem (theorem 1.2.2), this

implies that there exists an integer p0 such that the set�

Fp0 6= ;. Thus, there exists a point x0and a positive real number ↵ such that B(x0,↵) is included in Fp0 . Hence,

supn2N

kxk↵

kLn(x)kF supn2N

kxk↵

kLn(x+ x0)kF + kLn(x0)kF

supn2N

x2B(x0,↵)

kLn(x)kF + kLn(x0)kF

2p0.

Thus, for any x in E with a norm smaller than or equal to 1, we can write

kLn(x)kF ↵�1kLn(↵x)kF

2p0↵

·

Therefore, the sequence (Ln)n2N is bounded in L(E,F ). ⇤Back to the proof of theorem 3.3.2. Taking the limit, we get that

8x 2 B(0, 1) , kL(x)kF 2p0↵

·

We must now prove the upperbound on the norm of L. Let x be an element of E with anorm equal to 1. We can write that

kL(x)kF = limn!1

kLn(x)kF= lim inf

n!1kLn(x)kF

lim infn!1

kLnkL(E,F ) (because kxkE = 1).

All statements in the theorem are now proved. ⇤Theorem 3.3.1 is obtained by applying this result to the case F = K. ⇤

Although the following theorem has a very simple proof, it will be very useful.

Theorem 3.3.3. Let E and F be normed vector spaces. We consider an element A in L(E,F ),a sequence (`n)n2N in F

0, and an element ` in F

0. We then have

limf?n!1

`n = ` in F0 =) limf?

n!1

tA`n = t

A` in E0.

Proof. Indeed, for any x in E, we have

htA`n, xi = h`n, Axi.

58

We assume thatlimn!1

h`n, Axi = h`, Axi.

However, the definition of the transpose of a linear map yields h`, Axi = htA`, xi. So

limn!1

htA`n, xi = htA`, xi.

The theorem is proved. ⇤Proposition 3.3.1. Let (xn)n2N and (`n)n2N be two sequences in E and E

0respectively,

where E is a Banach space. Then,

⇣limf?n!1

`n = ` and limn!1

kxn � xk = 0⌘=) lim

n!1h`n, xni = h`, xi.

Proof. We begin by writing

h`n, xni = h`n, xn � xi+ h`n � `, xi+ h`, xi.

By the Banach-Steinhaus theorem, theorem 3.3.1, the sequence (`n)n2N is bounded, so thereexists a positive number M such that

|h`n, xni � h`, xi| Mkxn � xk+ |h`n � `, xi|.

Let " be a positive real number. There exists an integer n0 such that

n � n0 =) kxn � xk <"

2M + 1·

The point x is fixed, so there exists an integer n1 such that

n � n1 =) |h`n � `, xi| < "

2·

So, we haven � max{n0, n1} =) |h`n, xni � h`, xi| < ".


Once we have identified a dual space with another, we can define a notion of weak-?convergence on the latter as follows.

Definition 3.3.2. Let E and F be two Banach spaces, and let B be a bilinear form on F ⇥E

which identifies E0with F . By proxy, we say that a sequence (yn)n2N is weak-? convergent

to an element y in F if and only if the sequence (�B(yn))n2N is weak-? convergent to �B(y),which is stated mathematically like so:

8x 2 E , limn!1

B(yn, x) = B(y, x).

Let us illustrate the extension of this notion. For a given q in ]1,+1[, we consider a se-quence (yp)p2N of elements of `q(N). By theorem 3.2.1, we identify the dual of `q(N) with `q0(N).The fact that "the sequence (yp)p2N is weak-? convergent to an element y in `q(N)" means

8x 2 `q0

(N) , limp!1

1X

n=0

�yp(n)� y(n)

�x(n) = 0.

The following theorem is a fundamental result, which can be interpreted as a weak-?compactness property for the unit ball of a dual space.

59

Theorem 3.3.4 (weak-? compactness). Let E be a separable normed space. We consider a

bounded sequence (`n)n2N in E0. There exists an element ` in E

0and an extraction function

from N to N such that

limf?n!1

` (n) = `.

Proof. Let (ap)p2N be a dense sequence of E. We start be extracting from (ap)p2N a linearlyindependent family of vectors, and we will work on the subspace generated by this. Moreprecisely, we define '(0) as the smallest integer p such that ap is non-zero. Then, set '(1)the smallest integer p such that ap does not belong to Ka'(0). Let us assume that we havebuilt ('(0), · · · ,'(p)) such that

8p0 < p , a'(p) 62 Vp0def= Span{a'(0), · · · a'(p0)}.

Then, we set '(p+ 1) to be the smallest integer q such that aq is not in Vp. We define

V =[

p2N

Vp = Span�a'(p) , p 2 N

.

The family (a'(p))p2N is an algebraic basis of the vector space V . Henceforth, we omit theextraction function ' in our notation, and the sequence (ap)p2N is assumed to be a family oflinearly independent vectors of E such that Span{ap , p 2 N} is dense.

We are now going to use Cantor’s diagonal argument to construct an extraction function such that

8j 2 N , limn!1

h` (n), aji = �j . (3.7)

The sequence (h`n, a0i)n2N is a bounded sequence in K, so there exists an element �0 in Kand an extraction function �0 from N to N such that

limn!1

h`�0(n), a0i = �0.

Assume that we have constructed a finite family (�j)0jm of increasing functions from Nto N and a finite sequence (�j)0jm such that, for every j m,

limn!1

h`�0�···��j(n), aji = �j .

The sequence (h`�0�···��m(n), am+1i)n2N is bounded in K, so there exists an increasing func-tion �m+1 from N to N and an element �m+1 in K such that

8j m+ 1 , limn!1

h`�0�···��m+1(n), aji = �j .

Set (n) = �0 � · · · � �n(n). This is an increasing function from N to N. Indeed, observe thatany increasing function µ from N to N satisfies µ(n) � n. As a result, if n < m, we have

(n) = �0 � · · · � �n(n) < �0 � · · · � �n � �n+1 � �m(m) = (m).

Moreover, for any integer j, we have, for any n strictly greater than j,

h` (n), aji = h`�0��1·��j(�j+1��n(n), aji.

60

Thus, as any subsequence of a convergent sequence is convergent, we have

8j 2 N , limn!1

h` (n), aji = �j .

The linearity implies that for any x in V the sequence�h` (n), xi

�n2N

converges. We will nowprove that, for any x in E, the sequence

�h` (n), xi

�n2N is a Cauchy sequence. Set x in E. As

the space V is dense in E,

8" > 0 , 9y" 2 V /kx� y"kE <"

4Mwith M

def= sup

n2Nk`nkE0 .

Now, we can write that��h` (n+p), xi � h` (n), xi

�� h` (n+p), xi � h` (n+p), y"i

��+��h` (n+p), y"i � h` (n), y"i

��

+��h` (n), aj"i � h` (n), xi

��

k` (n+p)kE0kx� y"kE +��h` (n+p), y"i � h` (n), y"i

��

+ k` (n)kE0ky" � xkE "

2+��h` (n+p), y"i � h` (n), y"i

��.

As the sequence�h` (n), yi

�n2N converges, it is also a Cauchy sequence. Thus,

9n" / 8n � n" , 8p 2 N ,��h` (n+p) � ` (n), y"i

�� < "

2·

We deduce that

8" > 0, 9n" / 8n � n" , 8p 2 N ,��h` (n+p), xi � h` (n), xi

�� < "

The sequence�h` (n), xi

�n2N is a Cauchy sequence, so it converges to a limit which we will

denote `(x). By uniqueness of the limit, the map ` defined here is linear (check this as anexercise). Moreover, we have ��`(x)

�� MkxkE .

This ends the proof of the theorem. ⇤In the setting of Definition 3.3.2, Theorem 3.3.4 can be stated as follows.

Theorem 3.3.5. Let E be a separable Banach space, and let F be a Banach space. Let B

be a bilinear form which identifies E0

with F . Then, for any bounded sequence (yn)n2N of

elements of F , there exists an element y in F and an extraction function � such that

8x 2 E , limn!1

B(y�(n), x) = B(y, x).

This is a very important theorem. It means that when a space F can be identified as adual space, then any bounded sequence has a subsequence that is convergent in the weakersense given by definition 3.3.2. This is a widely used property, and we will see an importantapplication of this in the next chapter, where we will prove a diagonalisation theorem in Hilbertspaces (theorem 4.4.2 on page 75).

61

Chapter 4

Hilbert spaces

Introduction

This chapter is dedicated to the study of what are named Hilbert spaces. These are Banachspaces which have Euclidean-type norms, in other words norms that are induced by an innerproduct. These extend to infinite dimension the notion of Euclidean spaces, which have finitedimension. The key advantage is that we can talk about geometry in these spaces. Forexample, we can explain what two orthogonal vectors (which may be functions) are. This isa fundamental point of functional analysis worth insisting on: an element of a Banach space(often a function) is truly viewed as a point, or vector, of a geometric space.

The first section presents the concept of Hilbert spaces. In particular, we explain preciselywhat it means that a norm is associated with an inner product.

The second section shows how this structure generalises the notion of Euclidean space tothe infinite dimension setting. For example,

• it is possible to define an orthogonal projection onto a closed convex subset;

• it is possible to extend the familiar concept of orthonormal bases in finite-dimensionEuclidean spaces to the case of infinite dimension spaces.

In the third section, we show that the dual space of a Hilbert space H can be identifiedwith H itself. We will see in chapter 6 how this provides a spectacularly simple proof of atheorem guaranteeing existence and uniqueness of solutions of a partial differential equation.We finally note that the phenomenon seen in section 3.2 does not take place in Hilbert spaces.

In the fourth and final section of this chapter, we introduce the adjoint of a continuousoperator from a Hilbert space to itself, as well as the trickier, new notion of compact operators.The main result of this section is the generalisation of the classic diagonalisation theorem forsymmetric operators in finite dimension. This theorem will be used in chapter 6.

4.1 Orthogonality

In the rest of the course, it is understood that, if K = R, � = �, while, if K = C, this is theusual conjugation operation. We first recall the definition of an inner product.

Definition 4.1.1. Let E and F be two K-vector spaces. An map ` from E to F is said to be

antilinear if, for any pair (x, y) in E ⇥ F , and any scalar � in K, we have

`(�x+ y) = �`(x) + `(y).

63

A map f from E ⇥E to K is said to be sesquilinear1

if it is linear with respect to the first

variable, and antilinear with respect to the second one. In other words,

8� 2 K , 8(x, y, z) 2 E3

f(�x+ y, z) = �f(x, y) + f(y, z)

f(z,�x+ y) = �f(z, x) + f(z, y).

A map f from E ⇥ E to K is an inner product if and only if it is sesquilinear, Hermitian

and positive definite, i.e. it satisfies

8(x, y) 2 E2, f(y, x) = f(x, y) (Hermitian)

8x 2 E , f(x, x) 2 R+, (positive)

f(x, x) = 0 () x = 0 (definite).

Remark. An inner product is often denoted (·|·), and we write k · k2 = (·|·).

Proposition 4.1.1. Let E be a K-vector space, and let (·|·) be an inner product on E. The

map x 7! (x|x) 12 is a norm on E, and we have the following properties.

8(x, y) 2 E2, |(x|y)| kxkkyk (Cauchy-Schwarz inequality).

8(x, y) 2 E2, kxk2 + kyk2 = 2

��x+ y

2

��2

+ 2

��x� y

2

��2

(Appolonius’s theorem).

Proof. Obviously, we have

0 ��(x|y)

x

kxk � kxky��2

.

By expanding the right-hand side, we get

0 |(x|y)|2 � 2<e⇣(x|y)(x|y)

⌘+ kxk2kyk2.

To prove that x 7! (x|x) 12 is a norm, it suffices to note that

(x+ y|x+ y) (x|x) + 2<e(x|y) + (y|y) (x|x) + 2(x|x)

12 (y|y)

12 + (y|y)

⇣(x|x)

12 + (y|y)

12

⌘2, and that

(�x|�x) ��kxk2.

To prove Appolonius’s theorem, we simply observe that

2

��x+ y

2

��2

+ 2

��x� y

2

��2

=1

2kx+ yk2 + 1

2kx� yk2

=1

2(kxk2 + 2<e(x|y) + kyk2 + kxk2 � 2<e(x|y) + kyk2)

= kxk2 + kyk2.

The proposition is proved. ⇤1from the Latin “sesqui”, meaning “one and a half”

64

Exercise 4.1.1. Let (E, k · k) be a K-vector space. Assume that

8(x, y) 2 E2, kxk2 + kyk2 = 2

��x+ y

2

��2

+ 2

��x� y

2

��2

·

Prove that there exists a sesquilinear map such that kxk2 = (x|x). We recommend starting

with K = R, which is easier, before dealing with the case K = C.

An inner product allows one to define the notion of orthogonal vectors. This concept,which we are familiar with in the plane and in 3-dimensional space, will be extremely richwith applications in infinite-dimension spaces, in particular function spaces.

Definition 4.1.2. Let E be a vector space endowed with an inner product, and let x and y

be two elements of E. The vectors x and y are said to be orthogonal (denoted x ? y) if and

only if (x|y) = 0.

We recall the highly famous Pythagoras equality:

x ? y =) kx+ yk2 = kxk2 + kyk2.

Proposition 4.1.2. Let E be a K-vector space endowed with an inner product, and let A be

a subset of E. Define

A? def

= {x 2 E/ 8a 2 A , x ? a}.The set A

?is a closed vector subspace of E.

Proof. The linear map La from E to K defined by x 7! (x|a) is continuous, so its kernel is aclosed vector subspace of E. It is clear that

A? =

\

a2A

kerLa,

so A? is a closed vector subspace of E, as it is an intersection of the same. This proves the

proposition. ⇤Exercise 4.1.2. Let E be a K vector space endowed with a Hermitian inner product, and

let A be a subset of E. Prove that A?= A

? = Span(A)?.

Definition 4.1.3. Let H be a K-vector space endowed with an inner product (·|·). The

space H is called a Hilbert space if and only if the space H is complete for the norm associated

with the inner product.

The Hilbert space structure is absolutely fundamental. Many simple methods and conceptsfrom geometry in Euclidean spaces with finite dimension will reappear here.

4.2 Properties of Hilbert spaces

Pythagoras’s equality is extended to Hilbert spaces as follows.

Theorem 4.2.1. Let (xn)n2N be a sequence of elements of a Hilbert space H such that,

for n 6= m, xn ? xm. Then the series

X

n

xn is convergent if and only if the series

X

n

kxnk2 is

convergent, and in that case, ��X

n2Nxn

��2=X

n2Nkxnk2

65

Proof. Let us set Sq =X

pq

xp. Pythagoras’s theorem yields

X

pq

kxpk2 = kSqk2. (4.1)

If the right-hand side of the above equality is assumed to be convergent, then it is boundedindependently of q, and the series

X

n

kxnk2 is convergent. Conversely, if the seriesX

n

kxnk2

is convergent, then

kSq+q0 � Sqk2 =q+q

0X

q+1

kxpk2

1X

q+1

kxpk2.

By taking the limit in equality (4.1), we finish the proof of the theorem. ⇤Exercise 4.2.1. Let (xn)n2N be a sequence of elements of a Hilbert space H such that

1X

0

kxnk2 < 1.

Assume that there exists an integer N0 such that, if |n�m| � N0, the vectors xn and xm are

orthogonal. Prove that the seriesP

nxn is convergent, and that there exists a constant C,

which only depends on N0, such that

��X

n2Nxn

��2 C

X

n2Nkxnk2.

Many remarkable properties of Hilbert spaces are based on the following projection theo-rem.

Theorem 4.2.2 (projection onto a closed convex subset). Let � be a closed convex subset of

a Hilbert space H. Then, for any point x in H, there exists a unique point in �, which we

denote p�(x) and call the projection of x on �, such that

kx� p�(x)k = inf�2�

kx� �k.

Proof. Recall Appolonius’s equality: if � and �0 are two points in �, then we have

1

2k� � �

0k2 = kx� �k2 + kx� �0k2 � 2

��x� � + �0

2

��2

· (4.2)

Note that, since the set � is convex, the point� + �

0

2is also in �. We first prove uniqueness.

Assume that there are two points �1 and �2 in � which both minimise kx�gk for g 2 �. Then,Appolonius’s equality yields

1

2k�1 � �2k2 = kx� �1k2 + kx� �2k2 � 2

��x� �1 + �2

2

��2

= 2d2 � 2

��x� �1 + �2

2

��2

2d2 � 2d2 = 0.

66

This provides uniqueness. Now we prove existence. By definition of the infimum, there existsa minimising sequence, that is a sequence (�n)n2N of elements of � such that

limn!1

kx� �nk = ddef= inf

�2�kx� �k.

We use relation (4.2) again to get that, for any pair of integers (n,m),

1

2k�n � �mk2 = kx� �nk2 + kx� �mk2 � 2

��x� �n + �m

2

��2

·

As � is convex, the midpoint between �n and �m is also in �, thus,

1

2k�n � �mk2 kx� �nk2 + kx� �mk2 � 2d2.

Given the definition of the sequence (�n)n2N, the above yields that it is a Cauchy sequence.As H is a Hilbert space, it is complete, and, � being a closed subset of H, it is also complete,so the sequence (�n)n2N converges to a point in �. This ends the proof of the theorem. ⇤

Proposition 4.2.1. Let � be a closed convex subset of a Hilbert space H, and let x and x0

be two points in H. We denote p�(x) the unique point in � such that d(x,�) = kx� p�(x)k.Then, for any � in �,

<e(x� p�(x)|p�(x)� �) � 0 and kp�(x)� p�(x0)k kx� x

0k.

Proof. To get the first inequality, we observe that, as � is convex, for any � in [0, 1] and any �in �,

��x��(1� �)p�(x) + ��

��2 =��x� p�(x) + �(p�(x)� �)

��2 � kx� p�(x)k2.

A quick computation yields that

8� 2 [0, 1] , 8� 2 � ��2<e(x� p�(x)|p�(x)� �) + �kp�(x)� �k2

�� 0

which proves the first inequality. To prove the second one, we use the first result on thepoints x and � = p�(x0), then on the points x

0 and � = p�(x). This yields

<e�x� p�(x)|p�(x)� p�(x

0)�

� 0

and <e�x0 � p�(x

0)|p�(x0)� p�(x)�

� 0.

Adding the two together, we get

<e�x� x

0 � (p�(x)� p�(x0))��p�(x)� p�(x

0)�� 0.

Using the Cauchy-Schwarz inequality, this implies

kp�(x)� p�(x0)k2 <e

�x� x

0��p�(x)� p�(x

0)� kx� x

0k kp�(x)� p�(x0)k,

which ends the proof of the theorem. ⇤

67

Nearly all properties in Hilbert spaces can be seen as corollaries of the above theorem 4.2.2.

Corollary 4.2.1. Let F be a closed vector subspace of a Hilbert space H. Then,

H = F � F?

and x = pF (x) + pF?(x).

Proof. Let x be an element of H. For any point f in F , and for any real number �, we have

kx� pF (x) + �fk2 = �2kfk2 + 2�<e(x� pF (x)|f) + kx� pF (x)k2.

By definition of the projection on F , we must have

kx� pF (x) + �fk2 � kx� pF (x)k2.

This implies that, for any f in F , we must have

<e(x� pF (x)|f) = 0.

If K = R, this is all we need in order to state that

x� pF (x) 2 F?.

If K = C however, we must repeat the above argument with if instead of f to reach thatconclusion. Hence, we have proved that H = F + F

?. But, if x 2 F \ F?, then (x|x) = 0,

thus x = 0. The corollary is proved. ⇤

Corollary 4.2.2. Let A be a subset of a Hilbert space H. This subset A is total (i.e. the

vector subspace it spans is dense) if and only if A? = {0}.

Proof. We use the result of exercise 4.1.2, which states that A? = (Span(A))?. If A? = {0},

then Span(A) is equal to H, by Corollary 4.2.1, which precisely means that the subset A istotal.

Conversely, if A is total, then Span(A) is equal to H by definition, hence A? = {0}. The

property is proved. ⇤

Exercise 4.2.2. Let A be a subset of a Hilbert space H. Prove that

(A?)? = Span(A).

Definition 4.2.1. Let H be a separable Hilbert space with infinite dimension. We call an

orthonormal basis or Hilbertian basis (French term) of H, any sequence (en)n2N of elements

of H which is total, and such that

(en|em) = �n,m with �n,m = 1 if n = m and 0 otherwise. (4.3)

Remark. In this setting, an orthonormal basis is never a basis in the algebraic sense.

Exercise 4.2.3. Consider the space `2(N), and the sequence (en)n2N of elements of `

2(N)defined by

en(k) = �n,k.

Prove that (en)n2N is an orthonormal basis of `2(N). What is the vector space spanned by the

family (en)n2N?

Theorem 4.2.3. In every separable Hilbert space H, there exists an orthonormal basis.

68

Proof. Let (an)n2N be a total countable subset of H. We can assume that the family (an)n2Nconsists of linearly independent vectors. Indeed, we define the following extraction function

�(0) = min{n / an 6= 0} and �(n) = min�m/ am 62 Span{a0, . . . , a�(n)i

.

We let the reader check, as an exercise, that the vector subspace generated by the a�(n) is thesame as the one spanned by the an. We will now apply the Gram-Schmidt orthonormalisationmethod to this family. Let us recall the principle. First, we set

e1 =a1

ka1k·

We assume that the terms (ej)1jn satisfy property (4.3), so that

Span{a1, · · · , an} = Span{e1, · · · , en}.

Set

en+1 =fn+1

kfn+1kwith fn+1 = an+1 �

nX

j=1

(an+1|ej)ej .

It is quick to check the equalities in (4.3) are satisfied by en+1. The theorem is proved. ⇤

Theorem 4.2.4. Let H be a separable Hilbert space, and let (en)n2N be an orthonormal basis

of H. The map I defined by ⇢H �! `

2(N)x 7�! ((x|en))n2N

is a linear isometric bijection. In particular, the Bessel and Parseval equalities hold in H:

x =X

n2N(x|en)en and kxk2 =

X

n2N|(x|en)|2.

Proof. The main point to prove is that I is indeed a map from H to `2(N). We set

xqdef=X

pq

(x|ep)ep.

It is clear that

(x|xq) =X

pq

(x|ep)(x|ep)

=X

pq

|(x|ep)|2

= (xq|xq).

The Cauchy-Schwarz inequality states that kxqk2 kxk ⇥ kxqk. We deduce that, for anyinteger q, we have

qX

p=0

|(x|ep)|2 kxk2.

So I(x) does belong to `2(N).The fact that the map I is one to one is deduced from corollary 4.2.2, which states that the

orthogonal set of a total subset is the singleton {0}. That the map I is onto and an isometry,is deduced from theorem 4.2.1. Thus, theorem 4.2.3 is proved. ⇤

69

4.3 Duality in Hilbert spaces

Let H be a Hilbert space with finite dimension d; in other words, H is a Euclidean or Hermitianspace. We know that the map � defined by

�

⇢H �! H0

x 7�! �(x) : h 7! (h|x)

is an antilinear isometric bijection from H to H0. When H has infinite dimension, we will seefew changes.

Theorem 4.3.1 (Riesz representation theorem). Let H be a Hilbert space. We consider the

map � defined by

�

⇢H �! H0

x 7�! �(x) : h 7! (h|x)

This is an antilinear isometric bijection.

Proof. The fact that k�(x)kH0 kxk (and therefore �(x) is a continuous linear form) isobtained by using the Cauchy-Schwarz inequality. It is obvious that � is antilinear. Moreover,

h�(x), x

kxki = kxk ;

so � is an isometry. An isometry is always one to one, so it remains to prove that � is onto.To do this, let us consider an element ` in H0 \ {0}. Its kernel is a closed subspace of H, notequal to H. Let x be a non-zero vector in (ker `)?, such that h`, xi = kxk2. Both ` and �(x)have the same kernel, and there exists a non-zero h0 such that h`, h0i = h�(x), h0i, so �(x) = `.Thus, the theorem is proved. ⇤

The above theorem can be seen from a different angle when K = R, which will be usefulin chapter 5 to describe the duals of the L

p spaces.

Theorem 4.3.2. Let H be a real Hilbert space. For ` a continuous linear functional on H,

we define the function

F

(H �! Ru 7�! 1

2kuk2H � h`, ui.

The function F is bounded from below, and there exists a unique v in H such that F (v) =infu2H

F (u) which satisfies

8h 2 H , (v|h)H = h`, hi.

Proof. As the linear functional ` is continuous, we can write that

F (u) � 1

2kuk2H � k`kH0kuk

� 1

2(kukH � k`kH0)2 � 1

2k`k2

H0 .

Let µdef= inf

u2H

F (u), and let (un)n2N be a minimising sequence, that is a sequence of elementsof H such that

F (un) = µ+ "n with limn!1

"n = 0. (4.4)

70

By using Appolonius’s relation, we get��un � um

2

��2

H

=1

2kunk2H +

1

2kumk2H �

��un + um

2

��2

H

.

As ` is a linear map, we have��un � um

2

��2

H

=1

2kunk2H � h`, uni+

1

2kumk2H � h`, umi

� 2

✓1

2

��un + um

2

��2

H

� h`, un + um

2i◆.

Using (4.4), and by definition of µ, we obtain��un � um

2

��2

H

"n + "m.

This proves that the sequence (un)n2N is a Cauchy sequence, so it converges to an element v

in H. As the function F is continuous, we have F (v) = µ. Since, for any h in H and any realnumber t, we have

µ F (v + th) =1

2kvk2H + t(v|h) + 1

2t2khk2H � th`, hi,

from which we deduce

8h 2 H , 8t 2 R , t(v|h) + 1

2t2khk2H � th`, hi � 0,

which implies that8h 2 H , (v|h) = h`, hi.

Such an element v is of course unique, because if (v1�v2|h) = 0 for every h, we can choose h =v1 � v2, and this proves uniqueness. ⇤

We can deduce the following two consequences.

Definition 4.3.1. Let (xn)n2N be a sequence of elements of a Hilbert space H, and let x be an

element of H. The sequence (xn)n2N is said to be weakly convergent to x, denoted limfn!1

xn = x

or (xn)n2N * x, if and only if

8h 2 H , limn!1

(h|xn) = (h|x).

Observing this definition, we note that

limfn!1

xn = x () limfn!1

? �(xn) = �(x).

Among other things, the following theorem explains the terminology, why the convergenceis called weak.

Theorem 4.3.3. Let (xn)n2N be a sequence of elements of a Hilbert space H, and let x be

an element of H. Then, we have:

if limn!1

kxn � xk = 0, then limfn!1

xn = x ; (4.5)

if limfn!1

xn = x and limn!1

kxnk = kxk, then limn!1

kxn � xk = 0. (4.6)

Moreover, if (xn)n2N * x, then the sequence (xn)n2N is bounded.

71

Proof. The first point of the theorem is a direct consequence of the fact that

|(h|xn)� (h|x)| khk ⇥ kxn � xk.

For the second point, it suffices to write that

kxn � xk2 = kxnk2 � 2<e(x|xn) + kxk2;

As the sequence (xn)n2N is weakly convergent to x, we have

�2 limn!1

<e(x|xn) = �2kxk2.

This proves the second point. The third is a corollary of the Banach-Steinhaus theorem.Indeed, the weak convergence hypothesis means that, for any y in H,

limn!1

�(xn)(y) = limn!1

(y|xn) = (y|x).

Theorem 3.3.2 (Banach-Steinhaus) states that the sequence (�(xn))n2N is bounded in H. Bythe previous Riesz representation theorem, we get that (xn)n2N is a bounded sequence. Thetheorem is proved.

Example. Let H be a separable Hilbert space with infinite dimension. Consider an or-thonormal basis (en)n2N of H. By the Riesz representation theorem 4.3.1, for any x in H, thesequence ((x|en))n2N belongs to `2(N), so it converges to 0. However, the sequence (en)n2Ndoes not converge, as the norm of each en is equal to 1. This is therefore a sequence which isweakly convergent, but not convergent in norm.

Proposition 4.3.1. Let (xn)n2N and (yn)n2N be two sequences of elements in H such that

limn!1

xn = x and limfn!1

yn = y.

Then, limn!1

(xn|yn) = (x|y).

Proof. We simply write that

|(xn|yn)� (x|y)| |(xn � x|yn)|+ |(x|yn � y)| kxn � xk ⇥ kynk+ |(x|yn � y)|.

Theorem 4.3.3 yields that the sequence (yn)n2N is bounded, so

|(xn|yn)� (x|y)| Ckxn � xk+ |(x|yn � y)|,

The proposition is proved.

Exercise 4.3.1. Find two sequences (xn)n2N and (yn)n2N in H such that

limfn!1

xn = x , limfn!1

yn = y and limn!1

(xn|yn) 6= (x|y).

Exercise 4.3.2. Prove that, in a Hilbert space H, if a sequence (xn)n2N is weakly convergent

to x, then

kxk lim infn!1

kxnk

The following theorem is a weak compactness result for the unit ball of a Hilbert space.When the space does not have finite dimension, we know, by theorem 2.3.1, that the unit ballis not compact in the topology defined by the norm. However, we have the next theorem,which is merely theorem 3.3.4 on page 60 in this setting.

Theorem 4.3.4 (weak compactness). Let (xn)n2N be a bounded sequence in a separable

Hilbert space H. We can extract from (xn)n2N a weakly convergent subsequence.

72

4.4 The adjoint of an operator, self-adjoint operators

The adjoint of a linear map is a well-known notion in finite dimension. Given an inner producton a vector space H with finite dimension d, the adjoint of A is denoted A

?, and defined bythe relation

(Ax|y) = (x|A?y).

It is common knowledge that, if (Ai,j)1i,jd is the matrix of the map A in an orthonormalbasis, then the matrix representing A

? in the same basis is such that A?

i,j= Aj,i. Moreover,

there is a classic linear algebra theorem which states that if A is self-adjoint (i.e. A = A?),

then it can be diagonalised in an orthonormal basis.In this section, we generalise these concepts and results in Hilbert spaces with infinite

dimension.

Theorem 4.4.1. Let A be a continuous linear operator on a Hilbert space H. There exists a

unique linear operator A?, which is continuous on H, such that

8(x, y) 2 H⇥H , (Ax|y) = (x|A?y).

Moreover, the map from L(H) to itself, defined by A 7! A?, is an antilinear isometry.

We extend from this theorem the following definition.

Definition 4.4.1. The map A?

given above is called the adjoint of the operator A. An

element A of L(H) is self-adjoint if and only if A = A?.

Proof of theorem 4.4.1. Uniqueness of the operator A? is a consequence of the fact that H? ={0}. Let LA be the map defined by

LA

⇢H �! H0

y 7�! LA(y) : x 7�! (Ax|y).

It is clear that the map LA is a continuous antilinear map from H to H0. Set A? def= �

�1 �LA.By definition of A?, we have

(x|A?y) = (LAy|x) = (Ax|y).

So, the operator A? satisfies the desired relation with respect to A. We also have

kAk = supkxk1,kyk1

|(Ax|y)|

= supkxk1,kyk1

|(x|A?y)|

= kA?k.

So, theorem 4.4.1 is proved.

Exercise 4.4.1. Let H be a real Hilbert space, and let ⌦ be an open subset of H. We define

the “gradient” operator by

grad

⇢C

1(⌦;R) �! C(⌦,H)g 7�! �

�1Dg

Let g be a C1

function from ⌦ to ⌦0, another open set of H, and let f be a C

1function from ⌦0

to R. Compute grad(f � g).

73

Proposition 4.4.1. Let A be a continuous linear map on a Hilbert space H. The following

properties hold.

The?

operation is involutory, i.e. A = A??

. We also have

(kerA)? = ImA? and kerA = (ImA?)?. (4.7)

Finally, if a sequence (xn)n2N is weakly convergent to x, then the sequence (Axn)n2N is weakly

convergent to Ax.

Proof. The first point is simple: since (Ax|y) = (x|A?y), we have (A?y|x) = (y|Ax). To provethe equalities in (4.7), write that

8y 2 H , (Ax|y) = 0 () 8y 2 H , (x|A?y) = 0.

Restating the above equivalence in terms of sets yields that kerA and (ImA?)? are equal.

The other relation is deduced from this by taking the orthogonal.Let (xn)n2N be a sequence that converges weakly to x. For any y in H, we have

(y|Axn) = (A?y|xn).

So, the weak convergence of the sequence (xn)n2N implies the weak convergence of the se-quence (Axn)n2N. ⇤

Proposition 4.4.2. Let A be a self-adjoint operator in L(H). Then,

kAkL(H) = supkuk=1

|(Au|u)|.

Proof. First, we note that,

kAkL(H) = supkuk=1kvk=1

<e(Au|v)⇣taking v

def=

Au

kAuk

⌘·

It is clear that Mdef= supkuk=1 |(Au|u)| is less than or equal to kAkL(H). Let u and v be two

vectors in H with norm 1. Because A is self-adjoint, we have

<e(Au|v) =1

4

�(A(u+ v)|u+ v)� (A(u� v)|u� v)

�

M

4(ku+ vk2 + ku� vk2).

As ku+ vk2 + ku� vk2 = 2(kuk2 + kvk2), and the norms of u and v are equal to 1, we get

<e(Au|v) M,

which proves the proposition. ⇤We are now going to study a generalisation of the diagonalisation theorem for self-adjoint

operators on finite-dimension spaces. For this, we require an extra hypothesis on the operator:a compactness hypothesis. So, we first introduce the notion of compact operators. This notionis defined more generally for linear maps between Banach spaces.

Definition 4.4.2. Let E and F be two Banach spaces. An element u in L(E,F ) is called

compact if and only if, for any bounded subset A of E, the closure of the set u(A) is compact

in F .

74

Example. If u(E) is a vector space with finite dimension, then the operator u is compact.Indeed, if A is a bounded subset of E, the image of A is a bounded subset of the space u(E),which has finite dimension, and the closure of such a set is compact.

Exercise 4.4.2. Let E be the vector space of functions f from [0, 1] to R such that

supx 6=x0

|f(x)� f(x0)||x� x0| 12

< 1

1) Prove that

kfkEdef= |f(0)|+ sup

x 6=x0

|f(x)� f(x0)||x� x0| 12

is a norm on E, and that (E, k · kE) is a Banach space.

2) Prove that the map

◆

⇢E �! C([0, 1],R)f 7�! f

is compact.

Exercise 4.4.3. Let (A(n,m))(n,m)2N2 be a sequence of real numbers such that

X

(n,m)2N2

A2n,m

is finite.

1) Prove that the relation

(Ax)(n)def=X

m2NA(n,m)x(m)

defines a continuous operator A from `2(N) to itself, and that kAkL(`2)

⇣ X

(n,m)2N2

A2n,m

⌘ 12.

2) Let AN be an operator defined by (ANx)(n) = 1{0,··· ,N}(n)(Ax)(n). Prove that

limN!1

kAN �AkL(`2) = 0.

3) Deduce that the operator A is compact.

We now state the diagonalisation theorem for compact, self-adjoint operators on a Hilbertspace.

Theorem 4.4.2. Let A be a compact, self-adjoint operator on a separable Hilbert space Hwith infinite dimension. There exists a sequence of real numbers (�j)j2N, such that (|�j |)j2Nis nonincreasing and converges to 0, such that:

• for any j such that �j is non-zero, �j is an eigenvalue of A, and Ej

def= kerA � �j Id

is a subspace of H with finite dimension; moreover, if �j and �j0 are distinct, their

corresponding eigenspaces are orthogonal;

• if Edef= Span

[

j2N�j 6=0

Ej , then kerA = E?;

• we have kAkL(H) = maxj2N

|�j |.

75

Before we prove this theorem, we recall and re-prove the diagonalisation theorem for self-adjoint operators on Euclidean (or Hermitian) spaces with finite dimension.

Theorem 4.4.3. Let A be a self-adjoint operator on a Hilbert space with finite dimension N .

There exists a finite sequence (�j)1jN and an orthonormal basis (ej)1jN such that, for

any j, ej is an eigenvector corresponding to the eigenvalue �j .

Proof. Consider the quadratic (or Hermitian) form associated with A, that is

QA

⇢H �! Rx 7�! (Ax|x)

This is a continuous function on H. We denote SH the unit sphere of H. As the space hasfinite dimension, corollary 1.3.2 on page 24, a consequence of Heine’s theorem 1.3.3 on page 23,yields that there exists a point xM in SH such that

|(AxM |xM )| = M0def= sup

x2SH|(Ax|x)| (4.8)

We aim to show that xM is an eigenvector of A, corresponding to the eigenvalue ±M0. Thisproof is general, it does not depend on whether the dimension of the space is finite or not. Letus state a lemma.

Lemma 4.4.1. Let H be a Hilbert space (with finite or infinite dimension) and A is continuous

self-adjoint operator. Let x0 be a vector in SH such that

|(Ax0|x0)| = M0def= sup

x2SH|(Ax|x)|

Then x0 is an eigenvector of A, corresponding to the eigenvalue M0 or �M0.

Proof. By switching A for �A if needed, we can assume that (Ax0|x0) = M0. Observe that,for any non-zero vector y in H,

✓A

⇣y

kyk

⌘��y

kyk

◆ M0,

which implies that

8y 2 H , F (y)def= M0kyk2 � (Ay|y) � 0. (4.9)

We can observe that the point x0 satisfies F (x0) = 0, and is a minimum of the function F .Therefore, the differential of F at the point x0 is equal to zero. Alternatively, we can prove (4.9)directly, seeing that

8� 2 R , 8h 2 H , M0kx0 + �hk2 ��A(x0 + �h)|x0 + �h

�� 0

As F (x0) = 0, by expanding the above, we get

8� 2 R , 8h 2 H , 2�<e�M0x0 �Ax0

��h�+ �

2�M0khk2 �A(h|h)

�� 0.

As the discriminant of a non-negative polynomial is non-positive, we get that

8h 2 H , <e�M0x0 �Ax0

��h�= 0

In the Hermitian setting, we need to use the above relation with ih instead of h to find that,for any h in H,

�M0x0 �Ax0

��h�= 0, which ends the proof of the lemma. ⇤

76

Back to the proof of theorem 4.4.3. We are now going to prove a very simple linear algebraresult, which also holds regardless of whether the dimension of the space is finite or not.

Lemma 4.4.2. Let A be a self-adjoint operator on a Hilbert space H. When two eigenvectors

of A are assoicated with distinct eigenvalues, they are orthogonal.

Proof. Let u and v be two eigenvalues corresponding to distinct eigenvalues � and �0. We have

�(u|v) = (Au|v) = (Av|u) = �0(u|v).

This yields (�� 0)(u|v) = 0, so the lemma is proved. ⇤

Back to the proof of theorem 4.4.3. We define

E±

0def= ker(A±M0 Id) and H1

def= (E+

0 � E�

0 )? = (E+

0 )? \ (E�

0 )?. (4.10)

Lemma 4.7 states, in particular, that

(A±M0 Id)(H1) ⇢ (E±

0 )? and therefore A(H1) ⇢ H1 (4.11)

We reiterate the process on H1, and find either that there exist one or two non-zero eigenval-ues ±M1, or that 0 is an eigenvalue. As the dimension of H is finite, the decreasing sequenceof subspaces Hj is finite, and theorem 4.4.3 is proved. ⇤Proof of theorem 4.4.2. We follow the same lines as above, but some complications arise dueto the space having infinite dimension. Here are the three main tricky points to prove, orderedby increasing difficulty:

• the fact that the eigenspaces corresponding to non-zero eigenvalues have finite dimension;

• the fact that the sequence of eigenvalues converges to 0;

• the existence of a point xM that attains the supremum M0.

For the first point, we need to prove that the unit ball of the eigenspace Ej is compact,which, by theorem 2.3.1 on page 38, ensures that the dimension of Ej is finite.

Let (xn)n2N be a bounded sequence of Ej . As the operator A is compact, we can extract asubsequence (x'(n))n2N such that the sequence (Ax'(n))n2N converges. But, we have Ax'(n) =�jx'(n), and �j is non-zero, hence the unit ball of Ej is compact.

In order to prove the second point, let us consider an infinite sequence (�j)j2N of distincteigenvalues, and let (ej)j2N be a sequence of vectors with norm 1, such that ej is an eigenvectorcorresponding to the eigenvalue �j . The sequence (ej)j2N is weakly convergent to 0 (theproof is left as an exercise), and we can extract a strongly convergent subsequence from this.As Aej = �jej , we have

limj!1

|�j | kejk = 0.

As the norm of ej is equal to 1 for every j, it is the sequence (�j)j2N that converges to 0.

Now we need to construct the sequence (�j)j2N. The crucial element is the followinglemma.

Lemma 4.4.3. Let H be a separable Hilbert space with infinite dimension, and let A be a

compact, self-adjoint operator on H. There exists a vector xM on the unit sphere of H such

that

|(AxM |xM )| = M0def= sup

kxk=1|(Ax|x)|.

Moreover, AxM = ±M0xM .

77

Proof. If M0 = 0, then by proposition 4.4.2, we have A = 0, and there is nothing to prove. Sonow let M0 be positive. As (Ax|x) is real, and switching A for �A if needed, we assume that

M0 = supkxk=1

(Ax|x).

By definition, there exists a sequence (xn)n2N of elements of the unit ball B such that

limn!1

(Axn|xn) = M0

As the operator A is compact, there exist a subsequence, still denoted (xn)n2N, and an ele-ment y in H such that

limn!1

Axn = y.

By using theorem 4.3.4, we can extract another subsequence and assume that there exists x

in H such thatlimfn!1

xn = x.

We aim to prove that Ax = y. As the operator A is self-adjoint, we have, for any z in H,

limn!1

(xn|Az) = (x|Az) = (Ax|z).

It is also clear thatlimn!1

(Axn|z) = (y|z).

As a result, for any z in H, we have (y|z) = (Ax|z), which implies that y = Ax. So, byproposition 4.3.1,

limn!1

(Axn|xn) = (Ax|x).

The supremum M0 is therefore attained at a point x which is, of course, non-zero, since M0 isstrictly positive. Moreover, if we had kxk < 1, we would have

✓A

x

kxk

��x

kxk

◆=

M0

kxk2 > M0,

which contradicts the fact that M0 is the maximum value. So kxk = 1, which means that thesequence (xn)n2N converges to x in norm. ⇤

We can apply lemma 4.4.3 and work in the space

H1def=⇣ker(A�M0 Id) + ker(A+M0 Id)

⌘?

= ker(A�M0 Id)? \ ker(A+M0 Id)

?.

We prove that H1 is invariant under A. It is a simple lemma.

Lemma 4.4.4. Let B be an element of L(H). Then (kerB)? is invariant under B?.

Proof. Let x 2 (kerB)? and y 2 kerB. We have

(y|B?x) = (By|x) = 0,

hence B?x is in the orthogonal space of kerB. The lemma is proved. ⇤

78

Back to the proof of theorem 4.4.2. If we apply the above lemma with B = A±M0 Id, we getthat

(A±M0 Id)�ker(A±M0 Id)

�? ⇢

�ker(A±M0 Id)

�?,

because A is self-adjoint. This immediately implies that

A�ker(A±M0 Id)

�? ⇢

�ker(A±M0 Id)

�?,

hence AH1 ⇢ H1. We now study the operator A restricted to the Hilbert space H1. Theoperator A|H1

is is a compact, self-adjoint operator on H1 (exercise). Set

M1def= sup

kxk=1x2H1

(Ax|x).

By the preceding study, if M1 6= 0, this supremum is a maximum attained for at least onevector with norm 1 in (ker(A�M1 Id) + ker(A+M1 Id))

?. Thus M1 < M0. This process isreiterated as follows. Set

Hj

def=�E0 + · · ·+ Ej�1

�? and sup

kuk=1u2Hj+1

|(Au|u)| = Mj+1.

If Mj+1 = 0, proposition 4.4.2 implies that Hj+1 = kerA, and the procedure ends. If Mj+1 ispositive, the procedure continues.

If the sequence (Mj)j2N is a sequence of positive real numbers, set

E? =

\

j2NHj .

This means that8j 2 N , sup

kuk=1u2E

?

|(Au|u)| |�j |,

hencesupkuk=1u2E

?

|(Au|u)| = 0,

which implies, using proposition 4.4.2, that A|E? = 0. The proof of theorem 4.4.2 is complete.⇤

79

Chapter 5

Lp

spaces

Introduction

This chapter is dedicated to the study of functions from X to K which are integrable withrespect to a positive measure µ, defined on a �-algebra B of X, when elevated to a power p.Two fundamental examples will be looked at closely.

First, we will study the case where X = N, the �-algebra B is the power set of N, and µ

is the counting measure, i.e. the measure of a subset of N is the number of elements (a finitenumber or infinity) the subset contains. Then the space of functions which are p-th powerintegrable is the set `p(N), which we have already worked with in chapters 2 and 3.

The other important example is where X = Rd or a subset of Rd, endowed with the Borel�-algebra and the Lebesgue measure.

The first section will recall, without proof, a certain number of fundamental results fromintegration theory that we will constantly use in this chapter. They must imperatively beremembered.

In the second section, we prove that the spaces of (equivalence classes of) p-th powerintegrable functions are Banach spaces. The fact that such functions are defined modulonegligible sets adds some difficulty compared to the `p(N) spaces studied in chapter 2 (seetheorem 2.1.1 on page 31 and its proof). Hölder’s inequality plays an important role in theproof. In this section, we will also show a criterion for belonging to a space of p-th powerintegrable functions based on an inequality on a collection of weighted means: this is theimportant lemma 5.2.2. The notion that such a collection of weighted means is characteristicto each function is at the heart of the notion of weak derivative that we will introduce inchapter 6 and, more generally, of the concept of distributions (chapter 8).

The third section consists of the proof of the theorem stating that the space of continuousfunctions with compact support is dense in the space of p-th power integrable functions,when X is an open subset of Rd.

The fourth section is dedicated to defining the convolution of two functions defined on Rd.This is a massively used operation in analysis, and provides “explicit” approximations of p-thpower integrable functions using smooth functions (i.e. infinitely differentiable) with compactsupport.

The fifth and final section in this chapter aims to identify the dual of the space of p-th powerintegrable functions as the space of p0-th power integrable functions, where p

0 is the Hölderconjugate of p. This has already been observed in chapter 3 (theorem 3.2.1 on page 54). Wewill only show the case where p is in the interval [1, 2] in this chapter.

81

5.1 Measure theory and definition of the Lp spaces

In this section, we are going to define the Lp spaces and recall the main, basic theorems of

measure theory. Let X be a set, endowed with a �-algebra B and a positive measure µ on B.As is standard, the measure is assumed to be �-finite, which means that X is the union ofcountably many sets with finite measure.

Definition 5.1.1. Let p be in the interval [1,+1[. We denote Lp(X, dµ) the space of (equiv-

alence classes modulo equality almost everywhere of) measurable functions f such that

Z

X

|f(x)|pdµ < 1.

If p = 1, we denote Lp(X, dµ) the space of (equivalence classes modulo equality almost

everywhere of) measurable functions such that

kfkL1

def= sup

�� / µ{x / |f(x)| > �} > 0

= inf

�M / µ{x / |f(x)| > M}

= 0.

Let us prove that the two quantities in the definition of kfkL1 are equal. First, we notethat for any pair (�,M) such that µ{x / |f(x)| > M} = 0 and µ{x / |f(x)| > �} > 0, we havethat � is strictly less than M . Hence,

sup�� / µ{x / |f(x)| > �} > 0

inf

�M / µ{x / |f(x)| > M}.

Now, let M1 be a real number strictly greater than sup�� / µ{x / |f(x)| > �} > 0

. By

definition of the upper bound, we have µ{x / |f(x)| > M1} = 0. Thus,

inf�M / µ{x / |f(x)| > M} sup

�� / µ{x / |f(x)| > �} > 0

.

We now recall the fundamental theorems of measure theory.

Theorem 5.1.1 (monotone convergence, Beppo Levi). Let (fn)n2N be a sequence of positive,

measurable functions on X. We assume that, for every x, the sequence (fn(x))n2N is non-

decreasing. Let f be the function defined by

f

(X �! R+ [ {+1}x 7�! lim

n!1fn(x).

Then, in the set R+ [ {+1} with its usual topology, we have

limn!1

Z

X

fndµ =

Z

X

fdµ

Lemma 5.1.1 (Fatou). Let (fn)n2N be a sequence of measurable functions from X to [0,+1].Then, we have Z

X

(lim infn!1

fn)dµ lim infn!1

Zfndµ.

The link between almost-everywhere dominated convergence and convergence in the senseof the L

p norm is described in the next two theorems.

82

Theorem 5.1.2. Let p be a number greater than or equal to 1. Consider a sequence (fn)n2Nof elements of L

p, and f a function such that

limn!1

Z

X

|f(x� fn(x)|pdµ(x) = 0.

Then, there exists an extraction function such that

8 a.e dµ(x) , limn!1

f (n)(x) = f(x).

Theorem 5.1.3 (dominated convergence, Lebesgue). Let p be a real number greater than or

equal to 1, and let (fn)n2N be a sequence of functions in Lp(X, dµ). If, for almost every x

in X, we have

limn!1

fn(x) = f(x),

and if there exists a function g in Lp(X, dµ) such that, for almost every x in X,

|fn(x)| g(x),

then

f 2 Lp

and limn!1

kfn � fkLp = 0.

Theorem 5.1.4 (differentiability of functions defined by Lebesgue integrals). Let (X,µ) be

a measured space, ⌦ be an open subset of Rd, and f be a measurable function from ⌦ ⇥ X

to K. If, for almost every x in X, the function

z 7�! f(z, x)

is differentiable on ⌦, and if, for every x in ⌦, the functions

x 7�! f(z, x) and x 7�! Df(z, x)

are integrable on X, and, finally, if for every x in X and every z in ⌦, we have

|Df(z, x)| g(x) with g 2 L1(X, dµ),

then the function

z 7�!Z

X

f(z, x)dµ(x)

is differentiable on ⌦, and we have the formula

D

Z

X

f(z, x)dµ(x) =

Z

X

Df(z, x)dµ(x).

Theorem 5.1.5 (Fubini). Let (X1, µ1) and (X2, µ2) be two measured spaces, and let F be a

positive, measurable function from X1 ⇥X2 to [0,+1]. Then,

Z

X1⇥X2

F (x1, x2)dµ1(x1)⌦ dµ2(x2) =

Z

X1

dµ1(x1)

Z

X2

F (x1, x2)dµ2(x2)

=

Z

X2

dµ2(x2)

Z

X1

F (x1, x2)dµ1(x1).

By applying this theorem, one can solve the following exercise.

83

Exercise 5.1.1. Let f be a measurable function on a measured space (X,µ). Show that

Z

X

|f(x)|pdµ(x) = p

Z +1

0�p�1

µ(|f | > �)d�.

Theorem 5.1.6 (Fubini-Tonelli). Let (X1, µ1) and (X2, µ2) be two measured spaces, and

let F be a measurable function on X1 ⇥X2. The following two statements are equivalent:

i) the function F belongs to L1(X1 ⇥X2, dµ1 ⌦ dµ2);

ii) for almost every point x1 in X1, the function F (x1, ·) belongs to L1(X2, dµ2),

Z

X2

F (x1, x2)dµ2(x2) 2 L1(X1, dµ1),

and

��Z

X2

F (x1, x2)dµ2(x2)

��L1(X1,dµ1)

kFkL1(X1⇥X2,dµ1⌦dµ2).

Identically, i) is equivalent to: for almost any point x2 in X2, the function F (·, x2) belongs

to L1(X1, dµ1), Z

X1

F (x1, x2)dµ1(x1) 2 L1(X2, dµ2),

and

��Z

X1

F (x1, x2)dµ1(x1)

��L1(X2,dµ2)

kFkL1(X1⇥X2,dµ1⌦dµ2).

Moreover, if i) is satisfied, we have the formula

Z

X2

⇣Z

X1

F (x1, x2)dµ1(x1)⌘dµ(x2) =

Z

X1

⇣Z

X2

F (x1, x2)dµ2(x2)⌘dµ(x1)

=

Z

X1⇥X2

F (x1, x2)dµ1 ⌦ dµ2(x1, x2).

5.2 The Lp spaces are Banach spaces

The study of Lp spaces requires the introduction of the Hölder conjugate.

If p 2]1,1[ , p0 def=

p

p� 1, if p = 1, p

0 def= +1, and if p = +1 , p

0 def= 1.

The numbers p and p0 are called conjugate exponents, and, accepting the convention that

1

1 =

0, we have1

p+

1

p0= 1.

The first important new theorem of this chapter is the following.

Theorem 5.2.1. For any p in [1,1], (Lp(X, dµ), k · kLp) is a Banach space.

Proof. We start with the case where p is infinity. First of all, we note that if kfkL1 = 0, thereexists a sequence (�n)n2N which converges to 0 such that, for any n, the set {x / |f(x)| > �n} isnegligible. As the union of countably many negligible sets is also negligible, the set of points xsuch that f(x) is non-zero is negligible. Thus f is equal to 0 as an element of L1. The proofof the homogeneity of the L

1 norm is left as an exercise, and uses the fact that

{x / |�f(x)| > M} =nx / |f(x)| > M

|�|

o.

84

Now, let us consider f and g two (equivalence classes of) functions in L1, and set two positive

numbers Mf and Mg such that the measures of the sets {x / |f(x)| > Mf} and { / |g(x)| > Mg}are equal to zero. Note that

�x / |f(x) + g(x)| > Mf +Mg

⇢ {x / |f(x)| > Mf} [ {x / |g(x)| > Mg},

so the space L1 is stable under addition. Moreover, as the infimum of a set is a lower bound

of the set, we have kf + gkL1 Mf +Mg. As the infimum is the greatest lower bound, wehave kf + gkL1 kfkL1 + kgkL1 , thus we conclude that (L1(X, dµ), k · kL1) is a normedvector space.

In order to prove that (L1, k · kL1) is a Banach space, we are going to apply proposi-

tion 2.1.3 on page 29. Let us consider a Cauchy sequence (fn)n2N in (L1, k · kL1), and define

the set Em,p,n as follows:

Em,p,n

def=nx 2 X / |fm(x)� fm+p(x)| >

1

n+ 1

o.

Here, the equivalence classes of the functions fn and representatives of these classes are denotedthe same way. As the sequence (fn)n2N is a Cauchy sequence, we can define a function � from Nto N by

�(n) = minnm � �(n� 1) + 1 / 8m0 � m, 8p � 0 , kfm0 � fm0+pkL1 <

1

n+ 1

o·

By definition of the essential supremum, for any m � �(n), the measure of the set Em,p,n iszero. Thus, the set E defined as

Edef=

[

(n,p)2N2

m��(n)

Em,p,n

has zero measure. We can therefore choose representatives of the (classes of) functions fn thatare equal to zero on the negligible set E (these representatives are still denoted fn). Thus,

8n , 8m � �(n) , 8p , 8x 2 X , |fm(x)� fm+p(x)| <1

n+ 1

and the sequence (fn)n2N is a Cauchy sequence in B(X,K). We can now use proposition 2.1.3on page 29 to conclude that L

1(X, dµ) is complete.Now, we study the case where p is finite. The first step is to prove that (Lp(X, dµ), k · kLp)

is a normed vector space, where

kfkLpdef=⇣Z

X

|f(x)|pdµ(x)⌘ 1

p.

This is clear for p = 1. When p belongs to ]1,1[, we first note that, since

|f(x) + g(x)|p 2p(|f(x)|p + |g(x)|p),

Lp(X, dµ) is a vector space. Let us show that it is a normed space. This relies on Hölder’s

inequality, which we state and prove.

Proposition 5.2.1 (Hölder’s inequality). Let (X,µ) be a measured space, f be a function

in Lp(X, dµ) and g be a function in L

p0

(X, dµ). Then, the function fg is in L1(X, dµ), and

we have Z

X

|f(x)g(x)|dµ(x) kfkLpkgkLp0 .

85

Proof. First, note that if p = 1 or p = 1, there is nothing to prove. Moreover, we canassume that kfkLp = kgk

Lp0 = 1, and that the functions f and g are non-negative. Using theconvexity inequality (2.3) on page 29, we have

f(x)g(x) = (f(x)p)1p (g(x)p

0

)1p0 1

pf(x)p +

1

p0g(x)p

0

,

which proves Hölder’s inequality. Indeed, we integrating the above, and we getZ

X

|fg|dµ 1

p

Z

X

|f |pdµ+1

p0

Z

X

|g|p0dµ

1

p+

1

p0= 1.

This ends the proof of the proposition. ⇤Back to the proof of theorem 5.2.1. As f + g belongs to L

p, Hölder’s inequality implies thatZ

X

|f + g|pdµ Z

X

|f ||f + g|p�1dµ+

Z

X

|g||f + g|p�1dµ

⇣Z

X

|f |pdµ⌘ 1

p⇣Z

X

|f + g|(p�1) pp�1dµ

⌘ 1p0

+⇣Z

X

|g|pdµ⌘ 1

p⇣Z

X

|f + g|(p�1) pp�1dµ

⌘ 1p0

✓⇣Z

X

|f |pdµ⌘ 1

p+⇣Z

X

|g|pdµ⌘ 1

p

◆⇣Z

X

|f + g|(p�1) pp�1dµ

⌘1� 1p.

This implies that kf + gkLp kfkLp +kgkLp , and therefore that Lp(X, dµ) is a normed space.

Now we will show that these spaces are complete. The proof relies on the following lemma.

Lemma 5.2.1. If E is a normed space such that, for any sequence (xn)n2N of elements of E,

we have

X

n2NkxnkE < 1 =) SN

def=

NX

n=0

xn is convergent,

then E is a Banach space.

Proof. Recall proposition 1.3.2 on page 19, which states that a Cauchy sequence that hasa cluster point is convergent. Let (an)n2N be a Cauchy sequence in E: we will extract aconvergent subsequence, and this will yield the result. Define a map � from N to N by �(0) = 0and

�(n+ 1)def= min

(m � �(n) + 1 , sup

p�0kam � am+pkE 1

(2 + n)2

)·

Thus, for any integer n, we have

ka�(n+1) � a�(n)kE 1

(1 + n)2·

Set xndef= a�(n+1) � a�(n). We have

X

n2NkxnkE

X

n2N

1

(1 + n)2< 1.

86

So, by hypothesis, the sequenceNX

n=0

xn = a�(N)�a0 is convergent. Proposition 1.3.2 on page 19

states that any Cauchy sequence that has a cluster point is convergent, so the lemma is proved.⇤Back to the proof of theorem 5.2.1. Let (fn)n2N be a sequence of elements of Lp such that

X

n2NkfnkLp < 1.

Define the functions

SN (x) =NX

n=0

fn(x) and S+N(x) =

NX

n=0

|fn(x)|.

For any integer N , we havekS+

NkLp

X

n2NkfnkLp ,

which means that8n 2 N ,

Z

X

S+N(x)pdµ(x)

⇣X

n2NkfnkLp

⌘p

.

So the sequence (S+N(x))N2N is non-decreasing. The monotone convergence theorem 5.1.1

yields thatZ

X

S+(x)pdµ(x)

⇣X

n2NkfnkLp

⌘p

with S+(x) =

X

n2N|fn(x)|.

The function S+ takes finite values almost everywhere, and belongs to L

p. Thus, for almostany x, the series with general term fn(x) converges in C. We denote S its sum; S is an elementof Lp, and we have

��S �NX

n=0

fn

��p

Lp=

Z

X

��S(x)�NX

n=0

fn(x)��p

dµ(x).

We know that��S(x)�

NX

n=0

fn(x)�� 2S+(x).

As S+ is in L

p, the dominated convergence theorem yields that

limN!1

��S �NX

n=0

fn

��p

Lp= 0.

The theorem is proved. ⇤

Corollary 5.2.1. Let (fn)n2N be a sequence that converges to f in L1(X, dµ). Then there

exists an extraction function � such that

8a.e. x 2 X , limn!1

f�(n)(x) = f(x).

87

We are now going to make a few remarks on the Hölder inequality. We begin with thefollowing corollary.

Corollary 5.2.2. Let p and q be two real numbers strictly greater than 1, such that

1

p+

1

q 1.

Then the mapping ⇢Lp ⇥ L

q �! Lr

(f, g) 7�! fg

is bilinear and continuous if

1

r=

1

p+

1

q·

Proof. It suffices to apply Hölder’s inequality with s = r/p. This yieldsZ

X

|f(x)g(x)|rdx kfkrLpkgkrLq ,

which means thatkfgkLr kfkLpkgkLq ,

and the corollary is proved. ⇤

Corollary 5.2.3. If X has finite measure, then Lp(X, dµ) ⇢ L

q(X, dµ) if p � q.

Proof. As the function 1 is clearly in the space L1(X, dµ)\L

1(X, dµ), it belongs to Lp(X, dµ)

for any p. We even have8p 2 [1,1] , k1kLp = µ(X)

1p .

Corollary 5.2.2 implies that

kfkLq µ(X)1s kfkLp with

1

s=

1

q� 1

p,

so the corollary is proved. ⇤

We stress again that Hölder’s inequality is fundamental. It is also optimal in the followingsense.

Lemma 5.2.2. Let (X,µ) be a measured space. Let f be a measurable function and p be an

element of [1,1]. Assume that

supkgk

Lp01g�0

Z

X

|f(x)|g(x)dµ(x) < +1. (5.1)

Then f belongs to Lp, and

kfkLp = supkgk

Lp01

��Z

X

f(x)g(x)dµ(x)

��.

88

Proof. We begin with the case p = 1. By choosing the constant function equal to 1 for g, wehave g bounded with an L

1 norm equal to 1, and, using inequality (5.1), we get thatZ

X

|f(x)|g(x)dµ(x) < 1.

Now, consider g the function defined by

g(x) =f(x)

|f(x)| if f(x) 6= 0 and 0 otherwise.

The function g is bounded and its norm is equal to 1. Moreover,Z

X

f(x)g(x)dµ(x) =

Z

X

|f(x)|dµ(x).

The lemma is proved in this case.Now, let p be a real number strictly greater than 1, and consider an increasing sequence

of sets with finite measure (Kn)n2N, such that the union of these sets is equal to X. Set

f+n (x) = 1Kn\(|f |n)|f | and gn(x) =

f+n (x)p�1

kf+n k

pp0

Lp

·

It is clear that the function fn is non-negative, and belongs to L1 \ L

1, so it belongs to Lq

for any q, and we have

kgnkp0

Lp0 =1

kf+n kp

Lp

Z

X

f+n (x)(p�1) p

p�1dµ(x) = 1.

By definition of functions fn and gn, we haveZ

X

|f(x)|1Kn\(fn)gn(x)dµ(x) =

Z

X

f+n (x)gn(x)dµ(x)

=

✓Z

X

f+n (x)pdµ(x)

◆kfnk

�pp0

Lp

= kf+n k

p�pp0

Lp

= kf+n kLp .

Therefore, Z

X

f+n (x)pdµ(x)

✓sup

kgkLp01g�0

Z

X

|f(x)|g(x)dµ(x)◆

p

.

The monotone convergence theorem applied to the non-decreasing sequence�(f+

n )p�n2N im-

mediately yields thatkfkLp sup

kgkLp01g�0

Z

X

f(x)g(x)dµ(x).

Thus, if hypothesis (5.1) holds, the function f belongs to Lp. Now, we assume that f belongs

to Lp. Then, if we set

g(x) =f(x)|f(x)|p�1

|f(x)|⇥ kfkpp0

Lp

,

89

we have

kgkp0

Lp0 =1

kfkpLp

Z

X

|f(x)|(p�1) pp�1dµ(x) = 1 and kfkLp =

Z

X

f(x)g(x)dµ(x).

So the lemma is proved for every p in [1,1[. In the case where p = +1, consider a positivenumber � such that µ(|f | � �) is positive. Set E�

def= (|f | � �), and let g0 be a non-negative

function in L1, with support in E�, and with integral equal to 1. Then, if

g(x) =f(x)

|f(x)|g0 ,

we get Z

X

|f(x)g(x)|dµ(x) =Z

X

|f(x)|g0(x)dµ(x) � �

Z

X

g0dµ(x) � � .

The lemma is proved. ⇤

5.3 Density in Lp spaces

Throughout this chapter, ⌦ will denote an open set of Rd, endowed with the usual Euclideanmetric, and µ will be a non-negative measure such that the measure of every compact set isfinite. Let us present the main result of this section.

We begin with a digression on the topology of open sets of Rd.

Theorem 5.3.1 (characterisation of compact subsets of an open set of Rd). Let A be a closed

subset of an open set ⌦ of Rd, such that there exists a positive number r such that A ⇢ [�r, r]d.

Then

A is compact () infx2A

d(x,Rd \ ⌦) > 0.

Proof. We start by assuming that A is compact. As ⌦ is open, then, for any x in A, d(x,Rd\⌦)is strictly positive. By exercice 1.1.4 on page 13, we know that the function that maps a pointto its distance to a given set is continuous. Corollary 1.3.2 states that its infimum is attained,so it is strictly positive.

Conversely, let A be a closed set in ⌦ such that there exists a positive number r suchthat A ⇢ [�r, r]d, and that satisfies

�def= inf

x2A

d(x,X \ ⌦) > 0.

In order to prove that A is compact, it suffices to prove that A is closed in Rd. As A is closedin ⌦, this means that there exists a set B that is closed in Rd such that A = B \⌦. Let ⌦�/2be the closed set defined by

⌦�/2def= {x / d(x,X \ ⌦) � �/2}.

We have the following relations between sets:

A = A \ ⌦�/2 = B \ ⌦ \ ⌦�/2 = B \ ⌦�/2.

The set B \ ⌦�/2 is closed in Rd as the intersection of two closed sets, so A is closed in Rd.The theorem is proved. ⇤

90

The following result will also be useful to us.

Theorem 5.3.2. Let ⌦ be an open set of Rd. There exists a sequence of compact sub-

sets (Kn)n2N such that

[

n2NKn = ⌦ , Kn ⇢

�

Kn+1 and 8K compact subset of ⌦ , 9n /K ⇢�

Kn .

Proof. Set

Kn

def= B(0, n) \

⇢x 2 Rd

/ d(x,Rd \ ⌦) � 1

n

�·

As ⌦ is open, for each x in ⌦, d(x,Rd \ ⌦) is positive. So, there exists an integer n suchthat x 2 Kn. Moreover,

Kn ⇢�

B(0, n+ 1) \⇢x 2 Rd

/ d(x,Rd \ ⌦) > 1

n

�⇢

�

Kn+1 .

Hence the first point of the theorem. To prove the second one, we notice that[

n2N

�

Kn+1 = ⌦,

since Kn ⇢�

Kn+1. This proves the theorem. ⇤

Definition 5.3.1. Let ⌦ be an open set of Rd. We call an exhaustion of ⌦ by compact sets

a sequence of compact subsets of ⌦ which satisfy the results of theorem 5.3.2 above.

We now return to the Lp spaces.

Theorem 5.3.3. Let p be a number in [1,+1[. The space Cc(⌦), containing all continuous

functions with compact support in ⌦, is dense in Lp(⌦, dµ).

The proof of this theorem is long and intricate. We show it here for the reader’s culture.To prove this theorem, we will first show two density results which hold for functions defined on a

generic measured space. We start with the following proposition.

Proposition 5.3.1. If p belongs to [1,1[, then L1(X, dµ) \ L

1(X, dµ) is dense in Lp(X, dµ).

Proof. Consider a non-decreasing sequence (Kn)n2N of sets with finite measure, the union of which isequal to X. Set

fn = 1Kn\(|f |n)f.

It is clear that, for almost every x, we have

limn!1

fn(x) = f(x).

Moreover, we have the inequality

|fn(x)� f(x)|p 2p|f(x)|p.

The dominated convergence theorem then ensures that the result holds. ⇤Now, we prove a second density result.

Proposition 5.3.2. If p belongs to [1,1[, then the set of integrable simple functions (i.e. which onlytake a finite number of non-zero values on sets with finite measure) is dense in L

p(X, dµ).

91

Proof. We will use the following fact, which the reader can prove as an exercise: if (Y, d) is a metricspace, if A is a dense subset of Y , and if every element of A is the limit of a sequence of elementsof B, then B is also dense in Y . Courtesy of this, it suffices to show that every function in L

1(X, dµ)\L1(X, dµ) is the limit of a sequence of integrable simple functions. To do so, for a given function f

in L1(X, dµ) \ L

1(X, dµ), consider the sequence of functions (fn)n2N defined by

fndef=

+1X

j=0

j

n+ 11f�1([ j

n+1 ,j+1n+1 [)

+�1X

j=�1

j + 1

n+ 11f�1([ j

n+1 ,j+1n+1 [)

. (5.2)

It is obvious that8x 2 X , |fn(x)� f(x)| 1

n+ 1and |fn(x)| |f(x)|.

The dominated convergence theorem ensures that fn ! f in Lp, which means that there exists n0

such that, for n greater than n0,

kfn � fkLp <"

2· (5.3)

Moreover, as the function f is essentially bounded, the sums that appear in (5.2) are finite, since thereexists a real number � that satisfies

|j|n+ 1

� � =) µ

✓f�1⇣h

j

n+ 1,j + 1

n+ 1

h⌘◆= 0.

Hence each function fn only takes a finite number of values. Also, as the function f is integrable, the

sets f�1

✓j

n+ 1,j + 1

n+ 1

◆have finite measure, which ends the proof of the proposition. ⇤

Proof of theorem 5.3.3. Let f be a function in Lp(⌦). We must prove that, for every positive ", there

exists a continuous function g with compact support such that

kf � gkLp < ". (5.4)

By proposition 5.3.2, there exists an integrable simple function ef such that

kf � efkLp <"

2with ef =

NX

j=1

↵j1Aj , (5.5)

where the ↵j are real numbers, and the Aj are Borel sets with finite measure. We will accept thefollowing result at first.

Lemma 5.3.1. For any p in [1,1[, for any Borel set A with finite measure, and for any positivenumber ", there exists a continuous function h with compact support such that

kh� 1AkLp < ".

This allows us to state that, for each j in {1, · · · , N}, there exists a continuous function gj withcompact support in ⌦ such that

k1Aj � gjkLp "

2N |↵j |·

Now define the function g as follows: g def=

NX

j=1

↵jgj . We have

k ef � gkLp NX

j=1

|↵j | k1Aj � gjkLp

"

2·

Theorem 5.3.3 is therefore proved, providing we prove lemma 5.3.1. ⇤

92

Proof of lemma 5.3.1. Let (Kn)n2N be an exhaustion of ⌦ by compact sets (see theorem 5.3.2 anddefinition 5.3.1 on page 91). As A has finite measure, the characteristic function of A belongs to L

p

for every p. Moreover, for every x in ⌦, we have

limn!1

1Kn\A(x) = 1A(x) and 0 1Kn\A(x) 1A(x).

So, for every positive ", there exists n0 such that

k1Kn\A � 1AkLp <"

2· (5.6)

We now need a fundamental result from measure theory, called the regularity theorem for the Borelmeasure on Rd. We will use it without proof for now.

Theorem 5.3.4. For every set A belonging to the complete Borel �-algebra B with finite measure,

8" > 0 , 9 a compact set K , 9 an open set U / K ⇢ A ⇢ U and µ(U \K) < ". (5.7)

End of the proof of lemma 5.3.1. Thus there exists an open set U and a compact set K such that

K ⇢ Kn \A ⇢ U and µ(U \K) <⇣"

2

⌘p·

We can assume that the closure of U is compact, for example by replacing the open set U by U \Kn+B(0, "), which will still be denoted U . Set

h(x)def=

d(x, U c)

d(x, U c) + d(x,K)·

The function h is continuous, takes its values in [0, 1], and satisfies

h|X\U = 0 and h|K = 1.

We easily deduce that

|h� 1A| 1U\K and therefore |h� 1A|p 1U\K .

Hence,kh� 1AkLp <

"

2·

The lemma is now proved, because the function g is zero outside of the closure of U , which we knowto be compact. ⇤Proof of theorem 5.3.4. We present this proof for completeness and for the reader’s culture. The firstset consists of reducing the problem to the case where the open set ⌦ has finite measure. To do so,let us assume the theorem proved in that setting. Consider a subset A of the complete �-algebra of ⌦with finite measure. We then define the sequences

⌦p

def= ⌦ \B(0, p) , Ap

def= A \B(0, p) , B0

def= A0 and Bp = Ap \Ap�1 for p � 1.

The sets Bp satisfyp[

p0=0

Bp0 =p[

p0=0

Ap0 .

As the sets Bp are pairwise distinct, we have

1X

p=0

µ(Bp) = µ(A) < 1. (5.8)

93

Assertion (5.7) implies that

8" > 0 , 8p , 9Kp," compact , 9Up," open in ⌦ / Kp," ⇢ Ap ⇢ Up," and µ(Up," \Kp,") < "2�p�2.

Assertion (5.8), meanwhile, implies that the series with general term (µ(Kp,"))p2N is convergent. Sothere exists an integer p" such that

1X

p=p"+1

µ(Bp) <"

2· (5.9)

We then set

U"

def=

1[

p=0

Up," and K"

def=

p"[

p=0

Kp,"

We have

A =1[

p=0

Bp ⇢ U" and K" ⇢p"[

p=0

Bp ⇢ A.

Moreover, U" is an open subset of ⌦ as it is a union of open sets, and K" is compact as it is a finiteunion of compact sets. Finally, we have

U" \K" =1[

p=0

Up," \⇣ p"[

p=0

Kp,"

⌘c

⇢1[

p=0

Up," \p"\

p=0

Kc

p,"

⇢⇣ 1[

p=0

Up," \1\

p=0

Kc

p,"

⌘[

1[

p=p"+1

Kp,".

Since1[

p=0

Up," \1\

p=0

Kc

p,"⇢

1[

p=0

(Up," \Kp,"),

we deduce that

U" \K" ⇢⇣ 1[

p=0

(Up," \Kp,")⌘[

1[

p=p"+1

Kp,".

As Kp," is a subset of Bp, inequality (5.12) ensures that

µ(U" \K") 1X

p=0

µ(Up," \Kp,") +1X

p=p"+1

µ(Kp,")

1X

p=0

"2�p�2 +"

2

".

We must now prove the theorem in the case where ⌦ has finite measure. Let A be the set of subsetsof ⌦ which satisfy (5.7). We will prove that all open subsets belong to A, then that A is a �-algebra,which is enough to yield the result.

If A = U is open, then the (non-decreasing) sequence of compact sets (Kn)n2N defined by

KN

def=�x 2 U / kxk n and d(x, U c) � 2�n

satisfies8x 2 U , lim

n!11Kn(x) = 1U (x).

94

The monotone convergence theorem 5.1.1 implies that

limn!1

µ(U \Kn) = 0,

which ensures that U is an element of A. Now, let us consider an element A in A, " a positivenumber, U an open set, and K a compact set such that

µ(U \K) <"

2·

If C is a subset of B, we have the relation B \ C = B \ Cc, and therefore C

c \Bc = B \ C. Hence,

Uc ⇢ A

c ⇢ Kc and µ(Kc \ U c) = µ(U \K) <

"

2·

However, U c may not be compact. To circumvent this difficulty, we introduce an exhaustion of U c bycompact sets, that is a sequence Kn satisfying theorem 5.3.2 on page 91. It is clear that

8x 2 ⌦ , 1Kc\(Uc\Kn)(x) = 1Kc(x)� 1Uc\Kn(x) = 1Kc(x)� 1Uc(x)

The monotone convergence theorem yields that Ac belongs to A.

It remains to prove that, if (An)n2N is a sequence of elements of A, then the union of all the An

belongs to A. We start by proving the result for the union of a finite number of these. Let A1 and A2

be two elements of A. For any positive real number ", there exist two open sets U1 and U2, and twocompact sets K1 and K2, such that

pKj ⇢ Aj ⇢ Uj and µ(Uj \Kj) <

"

2·

K1 [K2 ⇢ A1 [A2 ⇢ U1 [ U2. (5.10)

The set U1 [ U2 is open, and the set K1 [K2 is compact. Moreover, we have

(U1 [ U2) \ (K1 [K2) = (U1 [ U2) \ (K1 [K2)c

=�U1 \ (K1 [K2)

c�[�U2 \ (K1 [K2)

c�

⇢ (U1 \Kc

1

�[ (U2 \K

c

2).

Hence

µ�(U1 [ U2) \ (K1 [K2)

� µ(U1 \K1) + µ(U2 \K2)

".

Using (5.8), we see that U1 [ U2 belongs to A. Also, as A is stable under taking the complements,if A1 and A2 belong to A, then

A1 \A2 = (Ac

1 [Ac

2)c 2 A.

So A is stable under finite union.Now, let (An)n2N be a sequence of elements of A. We follow a similar technique to the one we

used to reduce the proof to the case where ⌦ has finite measure. We define

Bn

def=

n[

k=0

Ak \k�1[

j=0

Ak.

The sets Bn are pairwise distinct, their union is equal to that of the An, and we haveX

n2Nµ(Bn) = µ

⇣[

n2NAn

⌘< 1.

95

As each set Bn belongs to A, there exists a sequence of open sets (Un)n2N and a sequence of compactsets (Kn)n2N such that

8n 2 N ,pKn," ⇢ Bn ⇢ Un," and µ(Un," \Kn,") < "2�n�2

. (5.11)

There exists an integer n" such that1X

n=ne+1

µ(Bn) "

2· (5.12)

We then set

U"

def=[

n2NUn," and K"

def=

n"[

n=0

Kn,"

The set U" is open, as it is the union of a sequence of open sets, and K" is compact as it is the unionof a finite sequence of compact sets. It is then obvious that

K" ⇢[

n2NAn ⇢ U". (5.13)

Moreover,

Kc

"=⇣ n"[

n=0

Kn,"

⌘c=

n"\

n=0

Kc

n,".

Thus, we can deduce that

U" \K" =⇣[

n2NUn,"

⌘\⇣ n"\

n=0

Kc

n,"

⌘

⇢✓⇣[

n2NUn,"

⌘\⇣\

n2NK

c

n,"

⌘◆[

1[

n=n"+1

Kn,"

⇢⇣[

n2NUn," \K

c

n,"

⌘[

1[

n=n"+1

Kn,"

⇢✓[

n2NUn," \Kn,"

◆[

1[

n=n"+1

Kn,".

By (5.11) and (5.12), we deduce that

µ(U" \K") X

n2Nµ(Un," \Kn,") +

1X

n=n"+1

µ(Kn,")

"

X

n2N2�n�2 +

1X

n=n"+1

µ(Bn)

".

Thus, the theorem is proved. ⇤From the above theorem, which can be admitted, we deduce the following fundamental result.

Corollary 5.3.1. Let p be a real number in the interval [1,1[. The space Lp(⌦) is separable.

Proof. Let (Kn)n2N be an exhaustion by compact sets of ⌦ (see definition 5.3.1 on page 91). Weonly need to find a countable subset A of Lp(⌦) such that, for any continuous function with compactsupport in Kn, we have

8" , 9g 2 A / kf � g||L1(Kn) ". (5.14)Let ( n)n2N be a sequence of continuous functions which take the value 1 near Kn, and which are sup-

ported in�Kn+1. The set A will be chosen as the set of functions written as nP , with P a polynomial

function with d variables and rational coefficients. Proving that this set satisfies property (5.14) is agreat exercise for the reader, which we highly recommend. ⇤

96

Exercise 5.3.1. Set x 2 Rd and ⌧x the mapping defined by

⌧x

⇢Lp �! L

p

f 7�! ⌧x(f) : y 7! f(x� y).

Prove that the map ⌧x is an isometry from Lp to L

p, and, if p is a real number, then

8f 2 Lp, lim

x!0k⌧xf � fkLp = 0 where f(y)

def= f(�y).

The following exercise shows how one can linearly “smooth” functions in Lp.

Exercise 5.3.2. 1) Let K be a compact subset of Rd. Prove that, for any positive ", there exists afinite sequence (Ak,")k2N of distinct sets with compact closure and diameter less than ", such that

1[

k=1

Ak," = K.

We consider the map Pn defined by

Pn

8><

>:

L1 �! L

1

f 7�! Pn(f)def=

1X

j=1

1

µ(Aj,

1n+1

)

✓Z

Aj, 1

n+1

f(x)dµ(x)

◆1A

j, 1n+1

.

2) Prove that Pn is a continuous linear map from L1(K) to L

1(K).3) Prove that, for any p, we have kPnkL(Lp;Lp) 1.4) Prove that, if f belongs to L

p and g belongs to Lp0, then

Z

K

f(x)Png(x)dµ(x) =

Z

K

Pnf(x)g(x)dµ(x) .

5) Prove that8f 2 L

p, lim

n!1kPnf � fkLp = 0 .

5.4 Convolution and smoothing

In this section, we work in the whole space Rd. The convolution operation is a critical one inthe study of functions on Rd.

Theorem 5.4.1. Let f and g be two functions in L1(Rd). Then, for almost every x in Rd

,

the function

y 7�! f(x� y)g(y)

is integrable, and the function F defined by

F (x)def=

Z

Rdf(x� y)g(y)dy

belongs to L1. It is called the convolution of f and g, and is denoted f ? g. The thus defined

? operation is a continuous bilinear map from L1 ⇥ L

1to L

1. The following properties hold:

kf ? gkL1 kfkL1kgkL1 and

Z

Rd(f ? g)(x)dx =

⇣Z

Rdf(x)dx

⌘⇣Z

Rdg(x)dx

⌘.

Proof. As the functions f and g are assumed to be in L1, we have

|f(x� y)|⇥ |g(y)| 2 L1(Rd ⇥ Rd) and

Z

Rd⇥Rd|f(x� y)|⇥ |g(y)|dxdy = kfkL1kgkL1 .

The conclusions of Fubini’s theorem 5.1.5 immediately yield the first part of the theorem. Thesecond part is a consequence of the Fubini-Tonelli theorem. ⇤

97

We can also define the convolution of a function in Lp and a function in L

p0 .

Theorem 5.4.2. Let f be a function in Lp, and g be a function in L

p0

. The formula

(f ? g)(x)def=

Z

Rdf(x� y)g(y)dy

defines a continuous bilinear map from Lp ⇥ L

p0

to L1

.

Proof. By Hölder’s inequality, for almost every x in Rd, the map

y 7�! f(x� y)g(y)

is integrable, and ��Z

Rdf(x� y)g(y)dy

�� kfkLpkgkLp0 .

The theorem is proved. ⇤Both of the above theorems extend to allow one to define the convolution of any two

functions, providing these are in the appropriate Lp spaces. More precisely, the following

holds.

Theorem 5.4.3 (Young’s inequality). Let (p, q, r) be a triple of real numbers such that

1 +1

r=

1

p+

1

q· (5.15)

Consider a couple of functions (f, g) in Lp ⇥ L

q. Then, the formula

(f ? g)(x)def=

Z

Rdf(x� y)g(y)dy

defines a function in Lr, and we have

kf ? gkLr kfkLpkgkLq .

Proof. This is an application of lemma 5.2.2. First, we assume that f and g are non-negative.Let ' be a positive function in L

r0 . We are going to bound

I(f, g,')def=

Z

Rd⇥Rdf(x� y)g(y)'(x)dxdy .

Let us assume that the inequality

I(f, g,') kfkLpkgkLqk'kLr0 (5.16)

is proved. This means that(x, y) 7�! f(x� y)g(y)'(x)

is in L1(Rd ⇥ Rd). Fubini’s theorem 5.1.5 states that, for almost every x,

y 7�! f(x� y)g(y)

98

is in L1(Rd

, dy). Thus, f ? g is a function defined for almost every x in Rd. Moreover, Fubini’stheorem ensures that

Z

Rdf ? g(x)'(x)dx =

Z

Rd

Z

Rdf(x� y)g(y)dy'(x)dx

I(f, g,') .

Inequality (5.16) and lemma 5.2.2 then imply that f ? g is in Lr, and that

kf ? gkLr kfkLpkgkLp .

If f and g do not have a specific sign (or are complex-valued), simply apply the above to |f |and |g|, and note that

��Z

Rdf(x� y)g(y)dy

�� Z

Rd|f(x� y)| |g(y)|dy.

Now we prove inequality (5.16). Without loss of generality, we can assume that kfkLp =kgkLq = 1. Let ↵ and � be two numbers in the interval ]0, 1[. We write that

I(f, g,') =

Z

Rd⇥Rdf(x� y)1�↵g(y)1��'(x)⇥ f(x� y)↵g(y)�dxdy .

We apply Hölder’s inequality with the measure

dµ(x, y) = f(x� y)↵g(y)�dxdy

and the pair (r, r0). As a result, we have

I(f, g,') I(1)(f, g,')

1r I

(2)(f, g,')1�1r

with I(1)(f, g,')

def=

Z

Rd⇥Rd|f(x� y)|(1�↵)r+↵|g(y)|(1��)r+�dxdy

and I(2)(f, g,')

def=

Z

Rd⇥Rd'(x)r

0

f(x� y)↵g(y)�dxdy .

We start by bounding I(1)(f, g,'). Fubini’s theorem for non-negative functions and the in-

variance of the Lebesgue measure under translations mean that

I(1)(f, g,') =

⇣Z

Rdf(x)(1�↵)r+↵dx

⌘⇣Z

Rdg(y)(1��)r+�dy

⌘

It is then natural to choose ↵ such that (1 � ↵)r + ↵ = p and � such that (1 � �)r + � = q,which yields

↵ =r � p

r � 1and � =

r � q

r � 1·

As kfkLp = kgkLq = 1, we get that I(1)(f, g,') = 1. Now we bound I

(2)(f, g). To do so,observe that, by the definitions of ↵ and �,

↵

p+�

q= 1.

99

Hölder’s inequality then implies that

Z

Rdf(x� y)↵|g(y)|�dy

✓Z

Rdf(x� y)↵⇥

p↵dy

◆↵p✓Z

Rdg(y)�⇥

q� dy

◆�q

kfk↵Lpkgk�Lq

1.

Hence,I(2)(f, g,') k'k

Lr0 .

Inequality (5.16), and therefore theorem 5.4.3, are proved. ⇤

Theorem 5.4.4. Let f and g be two functions in Lp

and Lq

respectively, such that

1

p+

1

q� 1.

Then

Supp (f ? g) ⇢ Adh (Supp f + Supp g).

Proof. Let x be a point in Rd, and ⇢ be a positive number such that

B(x, ⇢) \ (Supp f + Supp g) = ;.

For any bounded function ' which is zero outside B(x, ⇢), we haveZ

Rd⇥Rd'(x)f(x� y)g(y)dxdy =

Z

Rd⇥Rd'(x+ y)f(x)g(y)dxdy

= 0 .

The theorem is then proved by applying lemma 5.2.2 on X = B(x, ⇢). ⇤

Convolution is a crucial operation, because it is used to give an explicit means to smoothand approximate functions in L

p.

Theorem 5.4.5. Let ' be a function in L1(Rd) with integral equal to 1, and let p be a real

number greater or equal to 1. Set

'"(x) = "�d'

⇣x

"

⌘·

Then, for every function f in Lp,

lim"!0

k'" ? f � fkLp = 0.

Proof. Let f be a function in Lp, and ⌘ be a positive real number. There exist two continuous

functions g and with compact support such that

kf � gkLp <⌘

8k'kL1and k'� kL1 <

⌘

8kgkLp + 1·

100

As the integral of the function ', and therefore also that of '", is equal to 1, we can write that

'" ? f � f = '" ? f �⇣Z

Rd'"(x)dx

⌘f

= '" ? (f � g)�⇣Z

Rd'"(x)dx

⌘(f � g)

+ ('" � ") ? g �⇣Z

Rd('" � ")(x)dx

⌘g

+ " ? g �⇣Z

Rd "(x)dx

⌘g.

Seeing the choice of the function g we made, we have

k'" ? (f � g)kLp <⌘

8and

��⇣Z

Rd'"(x)dx

⌘(f � g)

��Lp

<⌘

8·

Also, we chose so that

k('" � ") ? gkLp <⌘

8and

��⇣Z

Rd('" � ")(x)dx

⌘g

��Lp

<⌘

8·

Ultimately, we get

k'" ? f � fkLp <⌘

2+

�� " ? g �⇣Z

Rd "(x)dx

⌘g

��Lp

.

But, by the definition of convolution, we gave, by a change of variables,

( " ? g)(x)�⇣Z

Rd "(x)dx

⌘g(x) =

Z

Rd (z)(g(x� "z)� g(x))dz .

We deduce immediately that��( " ? g)(x)�

⇣Z

Rd "(x)dx

⌘g(x)

�� Z

Rd| (z)| |g(x� "z)� g(x)|dz .

As the function g is uniformly continuous, for any ⌘, there exists a positive ↵ such that

|y � y0| ↵) |g(y)� g(y0)| < ⌘

2k kL1 | Supp g +B(0, "R)|1p + 1

·

SetR

def= sup

z2Supp

|z|.

Then we have" <

↵

R) k " ? g � gkL1 ⌘

2| Supp g +B(0, "R)|1p

·

Hence, we getk'" ? f � fkLp < ⌘.

Thus, the theorem is proved. ⇤

101

Important remark. As the reader will prove in the next exercise, this result does not holdwhen p = 1.

Exercise 5.4.1. Consider for f the Heaviside function H (the characteristic function of the set

of non-negative real numbers), and for ' any even continuous function with compact support

and integral equal to 1. Prove that k'" ? f � fkL1 � 1/2.

As we will see, the previous theorem is extremely important when it comes to smoothingfunctions. Indeed, theorem 5.1.4 on page 83 ensures that if ' is smooth with compact support,then the function '" ? f is also smooth with compact support if f is compactly supported.Thus, every function in L

p can be approximated by a compactly supported smooth function.First, the existence of such functions must be assured.

Proposition 5.4.1. Let f be the function from R to R defined by

f(x) = e1

x(x�1) if x 2 [0, 1] and f(x) = 0 otherwise.

This function is smooth with compact support.

Proof. It suffices to notice that

f(k)(x) =

Pk(x)

xk+1(1� x)k+1e

1x(x�1) ,

where Pk is a polynomial function with degree �k. We let the reader fill in the details.

Remark. The functions '" are known as mollifiers, or approximations to the identity.

Corollary 5.4.1. Let ⌦ be an open set of Rd. We denote D(⌦) the set of smooth functions

with compact support in ⌦. The space D(⌦) is not equal to {0}.

Proof. Consider the function '(x) = f(|x|), where f is given by proposition 5.4.1 above.Simply apply a homothety-translation to this function to fit its support in ⌦. ⇤

Corollary 5.4.2. For every real number p � 1, the space D(Rd) is dense in Lp(Rd).

Proof. By corollary 5.4.1, there exists a function ' that is smooth with compact support thathas its integral equal to 1. Consider the family ('")">0 defined by

'"(x) = "�d'

⇣x

"

⌘·

For any function f in Lp, we can find a continuous function g with compact support as close

to f as we like in Lp. Set

g" = '" ? g.

By proposition 5.4.4, the function g" has compact support. Also, as '" is smooth, so is g", bytheorem 5.1.4. The corollary is proved. ⇤

Corollary 5.4.3. Let ⌦ be an open set of Rd, and let p be a number in the interval [1,1[.

The space D(⌦) is dense in Lp(⌦).

102

Proof. Let (Kn)n2N be an exhaustion of ⌦ by compact sets (recall definition 5.3.1 on page 91).For f in L

p, set fndef= 1Knf . By theorem 5.1.3, we have

limn!1

kfn � fkLp(⌦) = 0.

Now set fn,"def= '" ? fn. This function is smooth, and its support is the compact set Kn +

Bf (0, "), which is the set of points that are at a distance at most " from Kn. By proposi-tion 5.3.1 on page 90, for " "n, the compact set Kn + Bf (0, ") is a subset of ⌦, hence thecorollary. ⇤

Corollary 5.4.4. Let K be a compact subset of ⌦, an open set in Rd. There exists a function

in D(⌦) such that = 1 on a neighbourhood of K.

Proof. Let � be a positive number such that, if

Kr

def= {x 2 ⌦ / d(x,K) 2r},

then K2� is a subset of ⌦ (such a number exists by theorem 5.3.1 on page 90). We consider asequence of mollifiers ('")" and set

"(x)def= '" ? 1K�(x) =

Z

K�

'"(x� y)dy .

There exists a constant C such that

x 62 K� +B(0, C") =) '" ? 1K�(x) = 0 .

So, if C" < �, then " 2 D(⌦), as K2� ⇢ ⌦. Finally, if x is in K��C" (which is non-emptysince � > C"), then, for any y 2 B(x,C"), y belongs to K�. Thus,

'" ? 1K�(x) =

Z

K�

'"(x� y)dy

=

Z

Rd'"(x� y)dy

= 1.

The corollary is proved. ⇤

5.5 Duality between Lp and L

p0

The main result of this section is the following theorem, which explains how continuous linearfunctionals on L

p spaces, when p is real, can be represented.

Theorem 5.5.1. Let p be in the interval [1,1[ and let B be the bilinear form defined by

B

8<

:

Lp0 ⇥ L

p �! K(g, f) 7�!

Z

X

f(x)g(x)dµ(x).

Then the bilinear form B identifies the dual of Lp(X, dµ) with L

p0

(X, dµ).

103

Before proving this theorem, we will show an important consequence of it, and make a fewcomments about the result.

Corollary 5.5.1. Let p be a number in [1,1[. Consider a bounded sequence (gn)n2N in Lp0

.

There exists an extraction function � and a function g in Lp0

such that

8f 2 Lp, lim

n!1

Z

X

g�(n)(x)f(x)dµ(x) =

Z

X

g(x)f(x)dµ(x). (5.17)

Proof. We apply theorem 3.3.4 on page 60 to the sequence (Ln)n2N of linear functionals on Lp

defined byhLn, fi

def=

Z

X

gn(x)f(x)dµ(x).

This yields the existence of an extraction function � and a linear form L such that

8f 2 Lp, lim

n!1

Z

X

g�(n)(x)f(x)dµ(x) = hL, fi.

Theorem 5.5.1 above then ensures the existence of a function g in Lp0 such that

8f 2 Lp, lim

n!1

Z

X

g(x)f(x)dµ(x) = hL, fi.

The corollary is proved. ⇤Let us consider the case where X is an open set in Rd, and µ is the Lebesgue measure.

Note that, as p is finite, the space Cc(⌦) is dense in Lp(⌦). We let the reader establish, as an

exercise, that assertion (5.17) is equivalent to the following:

8' 2 Cc(⌦) , limn!1

Z

X

g�(n)(x)'(x)dµ(x) =

Z

X

g(x)'(x)dµ(x). (5.18)

We are going to observe that this property is false when p0 is equal to 1, in other words when p

is infinity. Indeed, let � be a function in D(Rd) with integral equal to 1. As we showed whenproving theorem 5.4.5, if ("n)n2N is a sequence of positive numbers converging to 0, we have

8' 2 Cc(⌦), limn!1

Z

Rd

1

"dn

�

⇣x

"n

⌘'(x)dx = '(0).

But the linear functional �0 defined on Cc(⌦) by h�0,'i = '(0) is the Dirac measure at 0. Nofunction g in L

1 can represent this via the bilinear form B because, if such was the case,Z

⌦g(x)'(x)dx = '(0)

implies that g is zero except at x = 0, and therefore g is zero almost everywhere.

Proof of theorem 5.5.1 when p belongs to ]1, 2]. The theorem was proved in the case p = 2in chapter 4 on Hilbert spaces. The set X is assumed to be the union of countably manysubsets Kn, each with finite measure. Without loss of generality, we can assume that thesequence (Kn)n2N is non-decreasing, in the sense that Kn ⇢ Kn+1. Let � be a linear functionalon L

p(X, dµ), and let K be any subset of X with finite measure. We denote �K the restrictionof � to K, that is the linear form defined by

h�K , fi def= h�,1Kfi.

104

We are now going to work in the space Lp(K, dµ). Let � be a linear form on L

p(K, dµ).Assuming for now that there exists a unique function gK on K such that

8f 2 Lp(K, dµ) , h�, fi = B(gK , f) =

Z

K

gK(x)f(x)dµ(x), (5.19)

and such thatkgKk

Lp0 = k�Kk(Lp)0 , (5.20)we consider the non-decreasing sequence (Kn)n2N of subsets of X with finite measure such thatthe union of the Kn is the whole space X. Then there exists a sequence of functions (gn)n2Nsatisfying

8f 2 Lp(Kn, dµ) , h�Kn , fi = B(gn, f).

By lemma 5.2.2, we know that

gn 2 Lp0

and that kgnkLp0 = k�Knk(Lp)0 k�k(Lp)0 .

Moreover, the uniqueness of the function gn satisfying (5.19) and (5.20) ensures that, if m � n,

1Kngm = gn,

because the function 1Kngm also satisfies relations (5.19) and (5.20) for the linear form �Kn .The sequence (|gn|p

0

)n2N is therefore an non-decreasing sequence of functions such that

supn

Z

⌦|gn(x)|p

0

dµ(x) k�kp0

(Lp)0 .

The monotone convergence theorem ensures that the function

g(x) = limn!1

gn(x) 2 Lp0

and that kgkLp0 k�k(Lp)0 .

Lemma 5.2.2 then ensures that kgkLp0 = k�k(Lp)0 . Moreover, for any function f in L

p and forany integer n, we have

h�,1Knfi =Z

X

g(x)1Knf(x)dµ(x).

The dominated convergence theorem then finishes the proof off. But we are yet to proverelations (5.19) and (5.20). By Hölder’s inequality, we can say that

kfkLp =

✓Z

K

|f(x)|pdµ(x)◆ 1

p

✓Z

K

|f(x)|p2pdµ(x)

◆ p2⇥

1p

µ(K)1p(1�

p2 )

kfkL2µ(K)1p�

12 .

Thus, the linear form � appears as a continuous linear form on the space L2, so there exists

a function g in L2 such that

8f 2 L2, h�, fi =

Z

X

f(x)g(x)dµ(x).

We deduce that, for any function f in L2, we have

Z

K

|f(x)g(x)|dµ(x) k�k(Lp)0kfkLp .

But, by corollary 5.3.1, the space L1 \ L

p is dense in Lp. As p is between 1 and 2 here, and

since K is of finite measure, L2 is dense in Lp. By lemma 5.2.2, we deduce that g belongs

to Lp0 , and kgk

Lp0 = k�k(Lp)0 . Thus, (5.19) and (5.20) are proved.

105

Proof of theorem 5.5.1 when p belongs to ]2,1[. We start by proving that, for p in theinterval ]2,1[, the following inequality holds:

8(a, b) 2 R2,|a|p + |b|p

2��a+ b

2

��p

+p(p� 1)

2p�2

��a� b

2

��p

. (5.21)

As the function t 7! |t|p is twice continuously differentiable on R, the second-order Taylorformula with integral remainder between a and (a+ b)/2, then between b and (a+ b)/2, yields

|a|p =��a+ b

2

��p

+ pa� b

2

a+ b

2

��a+ b

2

��p�2

+ p(p� 1)⇣a� b

2

⌘2 Z 1

0(1� ✓)

��a+ b

2+ ✓

a� b

2

��p�2

d✓

and |b|p =��a+ b

2

��p

+ pb� a

2

a+ b

2

��a� b

2

��p�2

+ p(p� 1)⇣a� b

2

⌘2 Z 1

0(1� ✓)

��a+ b

2+ ✓

b� a

2

��p�2

d✓ .

The mean of these two equalities gives

|a|p + |b|p2

=��a+ b

2

��p

+p(p� 1)

2

⇣a� b

2

⌘2 Z 1

0(1� ✓)

✓��a+ b

2+ ✓

a� b

2

��p�2

+��b� a

2+ ✓

a+ b

2

��p�2◆d✓ .

If we have��a+ b

2+ ✓

a� b

2

�� 1

4|b� a| and

��a+ b

2+ ✓

b� a

2

�� 1

4|b� a|, (5.22)

then

✓|b� a| =��✓a� b

2� ✓

b� a

2

��

=��a+ b

2+ ✓

a� b

2� a+ b

2� ✓

b� a

2

��

��a+ b

2+ ✓

a� b

2

��+��a+ b

2� ✓

b� a

2

��

1

2|b� a|.

Assuming that a and b are distinct (there is nothing worth seeing otherwise), we get thatassertion (5.22) can only be satisfied if ✓ is less than or equal to 1/2. Hence,

|a|p + |b|p2

��a+ b

2

��p

+p(p� 1)

2

⇣b� a

2

⌘2��b� a

4

��p�2

Z 1

12

(1� ✓)d✓

��a+ b

2

��p

+p(p� 1)

2p�2

��b� a

2

��p

,

which proves inequality (5.21). The proof of the theorem will now continue, following the linesof that of theorem 4.3.2 on page 70. Let ` be a continuous linear form on L

p(X, dµ). Considerthe function

Fp

8<

:

Lp(X, dµ) �! R

f 7�! 1

pkfkp

Lp � h`, fi

106

The function Fp is bounded from below, because

Fp(f) �1

pkfkp

Lp � k`k(Lp)0kfkLp .

Set mp

def= inf

f2LpFp(f). We consider a minimising sequence (fn)n2N, that is a sequence such

thatFp(fn) = mp + "n with lim

n!1"n = 0.

We are going to prove that this is a Cauchy sequence. First, we apply inequality (5.21)with a = f(x) and b = g(x), then we integrate with respect to the measure dµ; we get that

8(f, g) 2�Lp(X, dµ)

�2, kfkp

Lp + kgkpLp � 2

��f + g

2

��p

Lp+

p(p� 1)

2p�1

��f � g

2

��p

Lp. (5.23)

Applying this inequality with f = fn and g = fm, we can write

1

pkfnkpLp +

1

pkfmkp

Lp � 2

p

��fn + fm

2

��p

Lp+

(p� 1)

2p�1

��fn � fm

2

��p

Lp.

After subtracting h`, fn + fmi, the inequality becomes

1

pkfnkpLp � h`, fni+ kfmkp

Lp � h`, fmi

� 2

✓��fn + fm

2

��p

Lp� h`, fn + fm

2i◆+

(p� 1)

2p�1

��fn � fm

2

��p

Lp.

By definition of Fp, this can be written as

(p� 1)

2p�1

��fn � fm

2

��p

Lp Fp(fn) + Fp(fm)� 2Fp

⇣fn + fm

2

⌘

As the sequence (fn)n2N minimises Fp, we deduce that

(p� 1)

2p�1

��fn � fm

2

��p

Lp "n + "m,

which ensures that (fn)n2N is a Cauchy sequence. Let f be its limit. As the function Fp iscontinuous, we deduce that Fp(f) = mp; in other words, the infimum of Fp is a minimum.

Now, for a given function h in Lp, consider the function gh from R to R defined by

gh(t)def= Fp(f + th).

Using the differentiability theorem for functions defined by integrals, we see that the function gh

is differentiable, and that

g0

h(t) =

Z

X

(f(x) + th(x))|f(x) + th(x)|p�2h(x)dµ(x)� h`, hi.

The point t = 0 is a minimum of the function gh, so its derivative vanishes at this point, whichmeans that

8h 2 Lp(X, dµ) ,

Z

X

f(x)|f(x)|p�2h(x)dµ(x) = h`, hi,

which ends the proof of Theorem 5.5.1. ⇤

107

Remark The result of Theorem 5.4.5 on families of mollifiers can be interpreted in terms ofduality: using the bilinear form

B(f, g) =

Z

⌦f(x)g(x)dx,

the space L1(⌦, dx) cannot be identified as the dual of a normed space containing continuous

functions with compact support in ⌦. Indeed, let ("n)n2N be a sequence of positive numbersconverging to 0. We consider a sequence ('"n)n2N of approximations of unity as defined inTheorem 5.4.5. Then, for any continuous function g with compact support, we have

limn!1

Z

⌦'"n(x)g(x)dx = g(0).

The "weak ? limit" of the sequence ('"n)n2N is not of the form B(f0, ·).

108

Chapter 6

The Dirichlet problem

Introduction

The aim of this chapter is to solve the Dirichlet problem on a bounded domain of Rd. Weshall see how this problem, when formulated in terms of functions of class C

1, will naturallylead to the introduction of the weak derivative, or quasi-derivative, concept. This concept willthen lead us to the more general sense of derivation for distributions.

In order to formulate the Dirichlet problem, we will need the following notations. Let ⌦ bea bounded open set of Rd. The set of continuously differentiable functions on ⌦ with supportin a compact subset of ⌦ is denoted C

10 (⌦). For any x in ⌦ and for any function u of class C1

on ⌦, we denoteru(x) = gradu(x) =

�@1u(x), · · · , @du(x)

�2 Rd

.

Moreover, we have

|ru(x)|2 =dX

j=1

|@ju(x)|2 and kruk2L2(⌦)

def=

Z

⌦|ru(x)|2dx =

dX

j=1

Z

⌦|@ju(x)|2dx.

If there is no risk of confusion, we will omet to write the dependence of these norms on ⌦.Let f be a function in L

2(⌦). We consider the functional

F

8><

>:

C10 (⌦) �! R

u 7�! 1

2kru(x)k2

L2 �Z

⌦f(x)u(x)dx =

1

2

dX

j=1

Z

⌦|@ju(x)|2dx�

Z

⌦f(x)u(x)dx

,

and we want to determine if F has a minimum, that is if there exists a function u in C10 (⌦)

such thatF (u) = inf

v2C10 (⌦)

F (v).

Note that F is a continuous function (exercise: prove it!).

In the first section of this chapter, we will present a classical approach to this problem, bystudying a minimising sequence from which we extract a subsequence that converges weaklyto a function u, and such that the sequences of partial derivatives is also weakly convergent.

This will lead us to introduce the concept of quasi-derivative of a function in L2 in the

second section. Originally used by the French mathematician Jean Leray in 1934 in order tosolve a global existence problem for solutions of the equation that governs the flow of viscous

109

incompressible fluids, it constitutes the first soundly defined departure from the classical senseof the derivative.

In the third section, this concept will allow us to define the Sobolev space, which is aHilbert space adapted for our problem. Once we have defined this space, solving the Dirichletproblem ��u = f , with u “vanishing on the boundary”, will be a simple application ofthe Riesz representation theorem (theorem 4.3.1 on page 70). This example illustrates thepower of functional analysis, by taking a seemingly complicated problem and solving it withelegant simplicity, providing the conceptual setting is chosen appropriately. As our final resultin this chapter, we will use the diagonalisation theorem for compact self-adjoint operators(theorem 4.4.2 on page 75) to “diagonalise” the Laplacian operator �.

6.1 A classical approach to the problem

We must first prove that the function F is bounded from below. This relies on the Poincaréinequality.

Proposition 6.1.1 (Poincaré inequality). Let ⌦ be a bounded open set of Rd. There exists

a constant C (which depends on ⌦) such that

8u 2 C10 (⌦) , kukL2 CkrukL2 .

Proof. Let u be a function in C10 (⌦). We first observe that the function equal to zero outside ⌦

and that coincides with u in ⌦ is of class C1 and has compact support in Rd. To simplify the

notations, this function will also be denoted u. We have

8(x1, x0) 2 ⌦ , u(x1, x0) =

Zx1

�1

@1u(y1, x0)dy1.

By the Cauchy-Schwarz inequality, this implies

8(x1, x0) 2 ⌦ , |u(x1, x0)|2 �(⌦)

Z +1

�1

|@1u(y1, x0)|2dy1.

Integrating this inequality with respect to x = (x1, x0), we getZ

⌦|u(x1, x0)|2dx �(⌦)2

Z

⌦|@1u(y1, x0)|2dy1dx0.

Thus, we have proved the Poincaré inequality. ⇤We deduce from this the following corollary.

Corollary 6.1.1. The functional F is bounded from below. Moreover, we have

8A > 0 , 9B / krukL2 � B =) F (u) � A. (6.1)

Proof. It suffices to say that, by the Poincaré inequality, we have

F (u) � 1

2kruk2

L2 � kfkL2kukL2

� 1

2kruk2

L2 � CkfkL2krukL2

� 1

2(krukL2 � CkfkL2)2 � 1

2C

2kfk2L2 .

110

So F is bounded from below. Moreover, if A is a positive number, by setting

B =�2A+ C

2kfk2L2

� 12 + CkfkL2 ,

we end the proof of the corollary. ⇤

Let us now consider a minimising sequence (un)n2N of functions in C10 (⌦), i.e. that satisfies

limn!1

F (un) = m = infv2C

10 (⌦)

F (v).

As we want to know if this infimum is reached, we are looking for a limit for this se-quence. Inequality (6.1) in Corollary 6.1.1 implies that the sequence (krunkL2)

n2N is bounded.The Poincaré inequality yields that the sequence (un)n2N is bounded in L

2. Thus, the se-quences (un)n2N and (@jun)n2N are all bounded in L

2. The weak compactness theorem 4.3.4on page 72 then states that there exist a function u and functions u(j) such that, after repeatedextracting of subsequences, we have

limfn!1

un = u and limfn!1

@jun = u(j)

.

Let us assume for a moment that u is in C10 (⌦), and let us consider ' in C

10 (⌦). By definition

of the weak convergence in a Hilbert space, we haveZ

⌦u(j)(x)'(x)dx = lim

n!1

Z

⌦@jun(x)'(x)dx.

The functions un and ' are functions in C10 (⌦), so we have, by integrating by parts,

Z

⌦u(j)(x)'(x)dx = � lim

n!1

Z

⌦un(x)@j'(x)dx.

Once again using the definition of weak convergence in L2, we deduce that

Z

⌦u(j)(x)'(x)dx = �

Z

⌦u(x)@j'(x)dx.

By Corollary 5.4.3 on page 102, we know that C10 (⌦) is dense in L

2(⌦), thus we have u(j) = @ju.

But there is no reason to believe that u is always of class C1; this is in fact generally false.However, the above computation shows us a link between the partial derivatives of u and thelimit functions u

(j). We will give mathematical foundations to this link in the next section.

6.2 The concept of quasi-derivatives

Here is what we are talking about.

Definition 6.2.1. Let u be a function in L2(⌦). We say that u has a partial quasi-derivative

in the direction j if and only if there exists a constant C such that

8' 2 C10 (⌦) ,

��Z

⌦u(x)@j'(x)dx

�� Ck'kL2

111

Fundamental remark. By theorem 5.3.1 on page 91, the space C10 (⌦) is dense in L

2(⌦).Theorem 1.2.4 on page 17, on prolonging continuous functions on a dense subspace, statesthat we can extend the linear form

`j

8<

:

C10 (⌦) �! R' 7�! �

Z

⌦u(x)@j'(x)dx.

to the whole L2 space. We denote ej this linear form. The Riesz representation theorem 4.3.1

on page 70 ensures that there exists a function u(j) in L

2 such that

8f 2 L2(⌦) ,

Z

⌦u(j)(x)f(x)dx = hej , fi.

In particular, for f = ' 2 C10 (⌦), this yields

8' 2 C10 (⌦) ,

Z

⌦u(j)(x)'(x)dx = �

Z

⌦u(x)@j'(x)dx.

We can thus give a more complete definition of the quasi-derivative.

Definition 6.2.2. Let u be a function in L2(⌦). We say that u has a partial quasi-derivative

in the direction j if and only if there exists a constant C such that

8' 2 C10 (⌦) ,

��Z

⌦u(x)@j'(x)dx

�� Ck'kL2 .

Moreover, the function u(j)

in L2(⌦) that satisfies

8' 2 C10 (⌦) ,

Z

⌦u(j)(x)'(x)dx = �

Z

⌦u(x)@j'(x)dx.

is called the quasi-derivative of u in the direction j.

To illustrate this notion, let us examine the following example.

Proposition 6.2.1. Let ⌦ =]� 1, 1[. We consider the function u(x) = |x|↵ with ↵ in ]1/2, 1].This function has a quasi-derivative, which is the function

x 7�! ↵sg(x)|x|↵�1.

Proof. We write that, for any C1 function ' with compact support in the interval ]� 1, 1[,

Z 1

�1|x|↵'0(x)dx = lim

"!0I"(') where I"(')

def=

Z�"

�1(�x)↵'0(x)dx+

Z 1

"

x↵'0(x)dx.

Integrating by parts, we get that

I"(') = "↵�'(�")� '(")

�+

Z�"

�1↵(�x)↵�1

'(x)dx�Z 1

"

↵x↵�1

'(x)dx

As ↵ is strictly greater than 1/2, the function |x|↵�1 is square-integrable. We can thus takethe limit in the above integrals, which yields

Z 1

�1|x|↵'(x)dx = �

Z 1

�1↵sg(x)|x|↵�1

'(x)dx.

The proposition is proved. ⇤The following proposition shows that we have effectively generalised the concept of deriva-

tives, at least when it comes to functions that are C1, continuously differentiable.

112

Proposition 6.2.2. If u belongs to C10 (⌦), then u has quasi-derivatives in all directions, and

these quasi-derivatives coincide with the partial derivatives.

Proof. By integrating by parts and using the Cauchy-Schwarz inequality, we have��Z

⌦u(x)@j'(x)dx

�� =��Z

⌦@ju(x)'(x)dx

��

k@jukL2k'kL2

Ck@jukL1k'kL2 .

Let u(j) be the quasi-derivative of u in the direction j. By definition, this function in L

2(⌦)satisfies

8' 2 C10 (⌦) ,

Z

⌦u(j)(x)'(x)dx = �

Z

⌦u(x)@j'(x)dx.

Moreover, by integrating by parts, we have

8' 2 C10 (⌦) ,

Z

⌦@ju(x)'(x)dx = �

Z

⌦u(x)@j'(x)dx.

This then implies that Z

⌦

⇣@ju� u

(j)⌘(x)'(x)dx = 0

As the space C10 (⌦) is dense in L

2(⌦), we conclude that u(j) = @ju, hence the proposition. ⇤

Notation. Henceforth, the quasi-derivative of u in the direction j will be denoted @ju.

6.3 The space H10(⌦) and the Dirichlet problem

In this section, we will define the appropriate function space for the minimisation problemintroduced at the beginning of the chapter.

Definition 6.3.1. Let ⌦ be an open set of Rd. We denote H

10 (⌦) the space of functions u

in L2(⌦) such that there exists a sequence (un)n2N of functions in C

10 (⌦) such that

limn!1

kun � ukL2(⌦) = 0

and such that the sequences (@jun)n2N are Cauchy sequences in L2(⌦).

The properties of this space are given in the following propositions.

Proposition 6.3.1. If u 2 H10 (⌦), then u has quasi-derivatives in all directions.

Proof. Let (un)n2N be a sequence of functions in C10 (⌦) that converges in L

2(⌦) to u, andsuch that the sequences (@jun)n2N are Cauchy sequences in L

2(⌦). As the space L2(⌦), there

exist functions u(j) in L

2(⌦) such that

limn!1

k@jun � u(j)kL2 = 0.

Integrating by parts, we can then write

�Z

⌦u(x)@j'(x)dx = � lim

n!1

Z

⌦un(x)@j'(x)dx

= limn!1

Z

⌦@jun(x)'(x)dx

=

Z

⌦u(j)(x)'(x)dx.


113

Proposition 6.3.2. Endowed with the following norm,

N(u)def=⇣kuk2

L2(⌦) +dX

j=1

k@juk2L2(⌦)

⌘ 12,

the space H10 (⌦) is a Hilbert space.

Proof. The fact that N is associated with the inner product

(u|v) def=

Z

⌦u(x)v(x)dx+

dX

j=1

Z

⌦@ju(x)@jv(x)dx

is an easy exercise we leave for the reader. Let us prove that the space H10 (⌦) is complete.

Consider a Cauchy sequence (un)n2N of elements of H10 (⌦). As the space L

2(⌦) is complete,there exist functions u and (u(j))1jd such that

limn!1

kun � ukL2(⌦) = limn!1

k@jun � u(j)kL2(⌦) = 0. (6.2)

By definition of the space H10 (⌦), for each n, there exists a function vn in C

10 (⌦) such that

kun � vnkL2(⌦) 1

nand k@jun � @jvnkL2(⌦)

1

n·

Thus, there exists a sequence (vn)n2N in C10 (⌦) such that

limn!1

kvn � ukL2(⌦) = limn!1

k@jvn � u(j)kL2(⌦) = 0.

So u 2 H10 (⌦), and u

(j) = @ju. By (6.2), we have

limn!1

N(un � u) = 0.

The proposition is proved. ⇤We are now going to examine the properties of the space H

10 (⌦) specific to the case where

the open set ⌦ is bounded.

Proposition 6.3.3. Endowed with the norm

kuk2H

10 (⌦)

def= kruk2

L2(⌦)def=⇣ dX

j=1

k@juk2L2(⌦)

⌘ 12,

the space H10 (⌦) is a Hilbert space.

Proof. Compared with proposition 6.3.2, the only thing we need to show is that k · kH

10 (⌦) is

a norm that is equivalent to N . To do so, we only need to prove that, for any u in H10 (⌦),

kukL2(⌦) CkrukL2(⌦).

Let (un)n2N be a sequence of C10 (⌦) such that

limn!1

kun � ukL2(⌦) = limn!1

k@jun � @jukL2(⌦) = 0

114

for every j in {1, · · · , d}. The Poincaré inequality implies that, for any n, we have

kunkL2(⌦) CkrunkL2(⌦).

Taking the limit, we get that, for any function u in H10 (⌦),

kukL2(⌦) CkrukL2(⌦).

⇤We have the following proposition, that the reader can prove as an exercise.

Proposition 6.3.4. The functional F extends continuously to the space H10 (⌦), and we have

m = infv2C

10 (⌦)

F (v) = infv2H

10 (⌦)

F (v)

This minimisation problem fits into the general, abstract setting: let H be a Hilbert space,and ` be a continuous linear form on H; can we find the minimum of the function F defined by

F`

(H �! Rv 7�! 1

2kvk2H � h`, vi

Formulated this way, the problem is immediately solved by applying theorem 4.3.2.

Remark. We return to the original problem. The unique minimum of the function F definedon H

10 (⌦) is the function u in H

10 (⌦) that satisfies

8v 2 H10 (⌦) ,

dX

j=1

Z

⌦@ju(x)@jv(x)dx = (u|v)

H10 (⌦) =

Z

⌦f(x)v(x)dx = (f |v)L2 . (6.3)

Let us assume that the solution u happens to be of class C2 on ⌦. Then, for any function '

of class C2 with compact support in ⌦, we have

dX

j=1

Z

⌦@ju(x)@j'(x)dx =

Z

⌦f(x)'(x)dx = (f |v)L2 .

Integrating by parts, we get

dX

j=1

Z

⌦@ju(x)@j'(x)dx =

Z

⌦f(x)'(x)dx = �

Z

⌦�u(x)'(x)dx.

When u is only in H10 (⌦), we will sat that it solves ��u = f “in the weak sense”, or “in the

distribution sense”.

The following compactness property, which will be proved in the next chapter, is crucialin proving the existence of a sequence of eigenvalues for the operator �.

Theorem 6.3.1 (Rellich). Any bounded subset of H10 (⌦) has compact closure in L

2(⌦).

We will now state a result that describes the structure of the Laplacian operator and itsaction on the space H

10 (⌦).

115

Theorem 6.3.2. There exist a non-decreasing sequence (�j)j2N of positive real numbers

that tends to infinity, and an orthonormal (Hilbertian) basis of the space L2(⌦), that we

denote (ej)j2N, such that the sequence (��1j

ej)j2N is an orthonormal basis of H10 (⌦), and such

that

��ej = �2jej "in the weak sense",

which means that, for any function of class C1

with compact support in ⌦, we have

dX

k=1

Z

⌦@kej(x)@k'(x)dx = �

2j

Z

⌦ej(x)'(x)dx.

Proof. Let us define an operator B by

B

⇢L2 �! H

10 (⌦) ⇢ L

2(⌦)f 7�! u

such that u is the solution of the Dirichlet problem in H10 (⌦). The operator B is continuous

from L2(⌦) to H

10 (⌦). By Rellich’s theorem, theorem 6.3.1 above, the operator B is compact

from L2(⌦) to H

10 (⌦). Let us prove that the operator B is self-adjoint. By (6.3), we have, for

any f in L2,

8v 2 H10 (⌦) , (Bf |v)

H10 (⌦) = (f |v)L2 .

By applying this relation to v = Bg, for some g in L2, we deduce that

(f |Bg)L2 = (Bf |Bg)H

10 (⌦) and therefore (f |Bg)L2 = (f |Bg)L2 .

Thus, the operator B is self-adjoint. Let us prove that it is one to one. Consider a function f

in L2 such that Bf = 0. By (6.3) again, this means that

8v 2 H10 (⌦) ,

Z

⌦f(x)v(x)dx = 0.

Corollary 5.4.2 states that D(⌦), and therefore H10 (⌦) which contains D(⌦), is dense in L

2(⌦).We then deduce that if Bf = 0, then

8g 2 L2,

Z

⌦f(x)v(x)dx = 0,

which implies that f = 0. Finally, we observe that if � is a non-zero eigenvalue, then it ispositive. Indeed, there exists in that case a non-zero function f� in L

2 such that

�kfk2L2 = (Bf |f)L2 = kfk2

H10.

Theorem 4.4.2 on page 75 ensure the existence of a non-increasing sequence (µj)j2N of numbersthat tends to 0 and an orthonormal basis (ej)j2N of L2(⌦) such that

Bej = µ2jej .

By applying the Laplacian, we get

ej = ��Bej = µ2j (��)ej ,

which means that��ej =

1

µ2j

ej .

To prove that the sequence (��1j

ej)j2N is an orthonormal basis of H10 (⌦), it suffices to observe

that�2j (ej |ek)L2 = (ej |ek)H1

0.

Theorem 6.3.2 is proved. ⇤

116

Chapter 7

The Fourier transform

Introduction

This chapter is fundamental, albeit brief. The Fourier transform is a very general operationon functions that are integrable with respect to the Haar measure (invariant by translation)on a locally compact commutative group G. Here, we will stick to the case of the space Rd

endowed with the Lebesgue measure.In the first section of this chapter, we define the Fourier transform of an integrable function,

and establish the main properties of this transform, which are:

• it turns differentiation into multiplication and vice-versa,

• it turns a convolution into a product.

Beyond these basic properties, we present a few examples in which we compute Fouriertransforms. The fundamental example of the Gaussian function is of particular note.

In the second section, we prove the Fourier inversion theorem which states that, whenthe Fourier transform of an integrable function is itself integrable, we can recover the functionfrom its Fourier transform. We deduce from this the Fourier-Plancherel theorem, which extendsthe Fourier transform to the space of square-integrable functions; this extension is, up to aconstant, an isometry of this space.

The third section applies these results in order to prove Rellich’s theorem used in theprevious chapter (see theorem 6.3.1 on page 115).

7.1 The Fourier transform on L1(Rd)

Definition 7.1.1. Let f be a function in L1(Rd). We call the Fourier transform of f , denoted bf

or F(f), the function defined by

F(f)

8<

:

Rd �! C⇠ 7�!

Z

Rde�ih⇠,xi

f(x)dxwhere h⇠, xi def

=dX

j=1

⇠jxj

Let us compute a few examples of Fourier transforms.

117

Proposition 7.1.1. Let a be a positive real number. We have the following results:

F(1[�a,a])(⇠) =2 sin a⇠

⇠

,

F�e�a|·|1R±

�(⇠) =

1

a± i⇠, so F

�e�a|·|

�(⇠) =

2a

a2 + ⇠2,

F⇣

a

a2 + | · |2⌘(⇠) = ⇡e

�a|⇠|,

and F�e�a|·|

2�(⇠) =

⇣⇡

a

⌘ d2e�

|⇠|2

4a .

Proof. By definition of the Fourier transform, we have

F(1[�a,a])(⇠) =

Za

�a

e�i⇠x

dx

= � 1

i⇠

�e�i⇠a � e

i⇠a�

=2 sin a⇠

⇠·

Again using the definition of the Fourier transform, we have

F�e�a|·|1R+

�(⇠) =

Z1

0e�(i⇠+a)x

dx

=1

a+ i⇠·

Likewise, we have F�e�a|·|1R+

�(⇠) =

1

a� i⇠· Adding these two results, we get

F�e�a|·|

�(⇠) =

2a

a2 + ⇠2·

Still using the definition of the Fourier transform, we have

F�e�a|·|

2�(⇠) = lim

R!1

ZR

�R

e�i⇠x�ax

2dx

= e�

⇠2

4a limR!1

ZR

�R

e�a(x+i

⇠2a)

2

dx.

The function z 7! e�az

2 is holomorphic on C, hence

�Z

R

�R

e�a(x+i

⇠2a)

2

dx�Z ⇠

2a

0e�a(R�iy)2

dy +

ZR

�R

e�ax

2dx+

Z ⇠2a

0e�a(R+iy)2

dy = 0

By taking the limit as R tends to infinity, we find thatZ

1

�1

e�a(x+i

⇠2a)

2

dx =

Z1

�1

e�ax

2dx

It is well known that Z

Re�ax

2dx =

⇣⇡

a

⌘ 12.

118

This proves the result in dimension 1. For the case in dimension d, it suffices to observe that

e�|x|

2=

dY

j=1

e�x

2j

and that, if fj are functions of L1(R), we have

F(f1 ⌦ · · ·⌦ fd) = bf1 ⌦ · · ·⌦ bfd,

which ends the proof of this proposition. ⇤

Note that the continuity theorem for functions defined by integrals yields that the Fouriertransform of an integrable function is continuous (and bounded) on Rd.

One of the main properties of the Fourier transform is that it turns differentiation intomultiplication and vice-versa, as shown by the following theorem.

Theorem 7.1.1. Let f be a function on Rdsuch that (1 + |x|)f(x) defines an integrable

function on Rd. Then the Fourier transform of f is of class C

1, and

@⇠jF(f) = �iF(Mjf) where (Mjf)(x)def= xjf(x). (7.1)

Let f be a function of class C1

such that f and its partial derivatives are integrable on Rd.

Then we have

F(@xjf) = iMjF(f). (7.2)

Proof. The two results rely on the fact that

@⇠je�ih⇠,xi = �ixje

�ih⇠,xi and @xje�ih⇠,xi = �i⇠je

�ih⇠,xi. (7.3)

In the first case, the hypothesis and the differentiability theorem for functions defined byintegrals ensure that f has partial derivatives, as well as relation (7.1). The second caserequires approximation. Let � be a C

1 function with compact support such that �(0) = 1,and let ("n)n2N be a sequence of positive numbers that converges to 0. Using (7.3), andintegrating by parts, we get

⇠jF(�("n·)f)(⇠) = �i

Z

Rd@xj

�e�ih⇠,xi

��("nx)f(x)dx

= �i

Z

Rde�ih⇠,xi

@xj

��("nx)f(x)

�dx.

Leibniz’s law and the dominated convergence theorem yield the second formula. ⇤

This theorem gives us another way to prove the formula for Gaussian functions in propo-sition 7.1.1.

Proof. By proposition 7.1.2 below, applied to the case where the linear map A is a homothety withration

pa, it suffices to show the result when a = 1. Let us assume first that the dimension d is equal

to 1, and set

g(⇠)def=

Z

Re�i⇠x

e�x

2

dx.

119

Relation (7.1) in theorem 7.1.1 implies that the function g is differentiable, and we have

g0(⇠) =

Z

R�ixe

�i⇠xe�x

2

dx

=

Z

Rie

�i⇠x1

2(e�x

2

)0dx.

Applying relation (7.2) from theorem 7.1.1 yields

g0(⇠) = �⇠

2g(⇠).

This differential equation is easy to solve: we get g(⇠) = g(0)e�⇠2

4 . A classical computation gives us

g(0) =

Z

Re�x

2

dx =p⇡.

So the result is proved in dimension 1. In general dimension d, we observe that

e�|x|2 =

dY

j=1

e�x

2j

and that, if the functions fj are in L1(R) ,we have

F(f1 ⌦ · · ·⌦ fd) = bf1 ⌦ · · ·⌦ bfd,


We will now provide a corollary which describes the properties of the Fourier transforms ofsmooth functions with compact support. Its proof (which we omit) merely consists of repeatedapplications of the above theorem.

Corollary 7.1.1. Let f be a function in D(Rd). Its Fourier transform bf is smooth, and

satisfies the following:

8N 2 N , 8↵ 2 Nd, sup⇠2Rd

(1 + |⇠|)N |@↵ bf(⇠)| < 1. where @↵ def= @

↵1x1

· · · @↵dxd.

We will now study the influence of linear transformations on the Fourier transform.

Proposition 7.1.2. Let f be a function in L1(Rd), and A be an invertible linear map from Rd

to Rd. We have

F(f �A)(⇠) = | detA|�1 bf � tA

�1.

Proof. We perform the change of variable y = Ax in the integral that defines the Fouriertransform; this yields

F(f �A)(⇠) = | detA|�1Z

Rde�i(⇠|A�1

x)f(x)dx.

By definition of the transpose, we have (⇠|A�1x) = (tA�1

⇠|x), hence the proposition. ⇤From theorem 7.1.1 and proposition 7.1.2 above, we will deduce the next proposition, which

shows how to compute the Fourier transform of a Gaussian function. This result will play acrucial role proving the inversion formula.

120

Proposition 7.1.3. Let a be a positive real number. We have

F⇣e�a|·|

2⌘(⇠) =

⇣⇡

a

⌘ d2e�

|⇠|2

4a .

The following proposition gives some ideas on how the Fourier transform works on convo-lutions.

Proposition 7.1.4. Let f and g be two functions in L1(Rd). We have

F(f ? g) = bf bg.

Proof. For every ⇠ in Rd, the function F⇠ from Rd ⇥ Rd to C defined by

F⇠(x, y)def= e

�ih⇠,xif(x� y)g(y) = e

�ih⇠,x�yif(x� y)e�ih⇠,yi

g(y)

is an element of L1(Rd ⇥ Rd). Fubini’s theorem ensures thatZ

Rde�ih⇠,xi

✓Z

Rdf(x� y)g(y)dy

◆dx =

Z

Rd

✓Z

Rde�ih⇠,x�yi

f(x� y)dx

◆e�ih⇠,xi

g(y)dy,

which shows precisely that the Fourier transform of a convolution is the product of the Fouriertransforms. ⇤

The Fourier transform is symmetric in the sense given by the following proposition.

Proposition 7.1.5. For any pair (f, g) of functions in L1(Rd), we have

Z

Rd

bf(x)g(x)dx =

Z

Rdf(⇠)bg(⇠)d⇠.

Proof. This is an immediate consequence of Fubini’s theorem applied to the integrable functionon Rd ⇥ Rd defined by (x, ⇠) 7�! f(⇠)e�ih⇠,xi

g(x). ⇤

7.2 The inversion formula and the Fourier-Plancherel theorem

We are now going to prove the fundamental theorem of Fourier analysis.

Theorem 7.2.1. Let f be a function in L1(Rd) such that its Fourier transform is also a

function in L1(Rd). Then, the function f is continuous, and we have

8x 2 R , f(x) = (2⇡)�d

Z

Rdeih⇠,xi bf(⇠)d⇠, (7.4)

as well as the Fourier-Plancherel relation

k bfk2L2 = (2⇡)dkfk2

L2 . (7.5)

Remarks.

• The result can be rewritten as

F2f = (2⇡)df where f(x)

def= f(�x). (7.6)

121

• This theorem means that a function f in S (defined in the next chapter) can be writtenas a superposition of oscillatory functions, the e

ih⇠,xi, and the Fourier transform appearsas the density of these oscillations.

Proof of theorem 7.2.1. By proposition 7.1.3, the theorem is already proved for Gaussianfunctions. Of course, we are going to use this result. Set

G(x)def= ⇡

�d2 e

�|x|2

and G"(x) =1

"dG

⇣x

"

⌘·

Proposition 7.1.3 states that bG"(⇠) = e�

"2|⇠|2

4 . By the dominated convergence theorem, wededuce from the fact that bf belongs to L

1 that

(2⇡)�d

Z

Rdeih⇠,xi bf(⇠)d⇠ = lim

"!0(2⇡)�d

Z

Rdeih⇠,xi bG"(⇠) bf(⇠)d⇠. (7.7)

As the function f belongs to L1, we know that bG"⌦f is a function in L

1(Rd⇥Rd). By Fubini’stheorem, we have

(2⇡)�d

Z

Rdeih⇠,xi bG"(⇠) bf(⇠)d⇠ = (2⇡)�d

Z

Rd⇥Rdeih⇠,xi bG"(⇠)e

�i(⇠|y)f(y)d⇠dy.

Proposition 7.1.3 applied with a = "2/4 states, in particular, that

(2⇡)�d

Z

Rdeih⇠,y�xi bG"(⇠)d⇠ = (2⇡)�d

⇣4⇡"2

⌘ d2e�

|x�y|2

"2 = G"(x� y).

Thus, we get that

8" > 0 , (G" ? f)(x) = (2⇡)�d

Z

Rdei(x|⇠) bG"(⇠) bf(⇠)d⇠.

Theorem 5.4.5 on page 100 states that if ("n)n2N is a sequence of positive real numbers con-verging to 0, then lim

n!1kG"n ? f � fkL1(Rd) = 0. Theorem 5.1.2 on page 83 adds that, up to

the extraction of a subsequence that we do not denote differently, for almost every x in Rd,we have

limn!1

(G"n ? f)(x) = f(x).

Statement (7.7) then ensures that, for almost every x in Rd,

f(x) = (2⇡)�d

Z

Rdeih⇠,xi bf(⇠)d⇠.

The continuity theorem for functions defined by integrals ends the proof of (7.4), as it yieldsthat the right-hand side of the above equality, and therefore f , is continuous.

To prove (7.5), first observe that the fact that f and bf are in L1(Rd) implies that f is

bounded, and thus both f and bf are square-integrable. Noting that F(f) = F(f) and byusing proposition 7.1.5, we can write that

(2⇡)�dk bfk2L2 =

Z

Rd(2⇡)�d(Ff)(⇠)(F(f)(⇠)d⇠

=

Z

Rd(2⇡)�d(F2

f)(x)f(�x)dx.

The Fourier inversion formula as stated in (7.6) then ensures that

(2⇡)�dk bfk2L2 =

Z

Rdf(�x)f(�x)dx = kfk2

L2 ,

which ends the proof of the fundamental theorem of Fourier analysis. ⇤

122

This theorem, at the base of Fourier analysis, has countless applications. Let us providetwo corollaries.

Corollary 7.2.1. The Fourier transform extends into a continuous, invertible linear map

from L2(Rd) to L

2(Rd), which satisfies

8f 2 L2(Rd) , k bfkL2(Rd) = (2⇡)

d2 kfkL2(Rd).

Proof. The Fourier transform is well defined on D(Rd), which, by corollary 5.4.2 on page 102,is a dense subspace of L

2(Rd). Corollary 7.1.1 allows us to apply the Fourier-Plancherelrelation (7.5) to functions in D(Rd). The extension theorem 1.2.4 on page 17 then allows usto extend F into a continuous linear map from L

2(Rd) to itself, and the Fourier-Plancherelrelation naturally extends to the whole L

2 space. The only point left to prove is that F isonto. Proposition 7.1.5 implies, in particular, that

8(f, g) 2 D(Rd) , (Ff |g)L2(Rd) = (f |Fg)L2(Rd). (7.8)

To prove that F is onto, we first observe that the Fourier-Plancherel equality impliesthat F(L2) is closed in L

2, then we prove that F(L2) is dense by using criterion 4.2.2 onpage 68, which states that we only need to show that (F(L2))? consists only of the zerofunction. Indeed, if g is a function in L

2(Rd) such that

8f 2 L2(Rd) , (F(f)|g)L2(Rd) = 0,

then relation (7.8), extended to the whole L2 space, implies that Fg is equal to zero, and thus

so is g, since F is one to one. ⇤

Corollary 7.2.2. Let f and g be two functions in L1(Rd) with Fourier transforms also

in L1(Rd). Then we have

F(fg) = (2⇡)�d bf ? bg.

Proof. We apply proposition 7.1.4 to bf and bg. This yields

F( bf ? bg) = F2fF2

g = (2⇡)2df g,

which can be rewritten as( bf ? bg) = (2⇡)2dF�1(f g).

The Fourier inversion formula in theorem 7.2.1 ensures the result. ⇤

7.3 Proof of Rellich’s theorem

The goal of this section is to prove the Rellich’s compactness theorem 6.3.1, which we recall.This proof is presented here for cultural purposes only.

Theorem 7.3.1. Any bounded subset of H10 (⌦) has compact closure in L

2(⌦).

Proof. We will prove that, for any positive number ↵, we can cover a bounded part A of H10 (⌦) with

a finite number of balls with radius ↵ in the L2 norm. To do so, we start by proving that if � is a

function in D(B(0, 1)) with integral equal to 1, then��1

"d�

⇣ ·"

⌘? a� a

��L2

C� "krakL2 (7.9)

123

where C� is a constant that depends on �. By the Fourier-Plancherel relation (theorem 7.2.1 onpage 121) and by theorem 8.2.5, we have

��1

"d�

⇣ ·"

⌘? a� a

��L2 = (2⇡)

d2

��F⇣ 1

"d�

⇣ ·"

⌘? a� a

⌘��L2

= (2⇡)d2

��b�("·)ba� bakL2 .

The mean value inequality and theorem 7.1.1 imply that

|b�(⇣)� 1| |⇣|kDb�kL1

|⇣| max1jd

kxj�kL1(Rd).

Thus, we deduce that ��1

"d�

⇣ ·"

⌘? a� a

��L2

C� "k | · |bakL2 .

Proposition 7.1.1 and the Fourier-Plancherel relation then yield inequality (7.9). We will now provethat, for any " in the interval ]0, 1[, the set

B"

def=n 1

"d�

⇣ ·"

⌘? a , a 2 BH

10 (⌦)(0, 1)

o

is a subset of C(⌦ + B(0, 1)) with compact closure. To do so, we first observe that, by the Poincaréinequality, we have, for any x in ⌦+B(0, 1),

��1

"d�

⇣ ·"

⌘? a(x)

�� 1

"d�

⇣ ·"

⌘��L2kakL2

C

"d2

k�kL2krakL2 . (7.10)

Moreover, by the mean value inequality, we have��1

"d�

⇣ ·"

⌘? a(x)� 1

"d�

⇣ ·"

⌘? a(x0)

�� |x� x0|��1

"dD

⇣�

⇣ ·"

⌘? a

⌘��L1

|x� x0| "� d

2 k�kL2krakL2 .

For any " in the interval ]0, 1[, the set B" is an equicontinuous subset of C(⌦ + B(0, 1)). By in-equality (7.10) and Ascoli’s theorem (see theorem 2.4.1 on page 40), the set B" has compact closurein C(⌦ + B(0, 1)). It can thus be covered by a finite number of balls (for the norm inducing uniformconvergence on ⌦+B(0, 1)) with radius

↵

�µ(⌦+B(0, 1))

� 12

·

As, for any bounded function f supported in ⌦+B(0, 1), we have

kfkL2 �µ(⌦+B(0, 1))

� 12 kfkL1 ,

we have proved that B" has compact closure in L2(⌦), owing to (7.9). ⇤

124

Chapter 8

Tempered distributions in dimension 1

Introduction

This chapter is one of the endgames of this course. As we saw in chapter 3, identifying aBanach space as the dual of another via a continuous bilinear form (see definition 3.2.1 onpage 54) is an important ability that allows one to extract subsequences from any boundedsequence that converge in a weaker sense; we called this sense weak-? convergence.

When we studied the spaces of p-th power integrable functions in chapter 5, we saw thatunderstanding families of weighted means,

⇣Z

⌦f(x)'(x)dx

⌘

'2A

,

e.g. with ' in a dense subset A, could sometimes be better illustrations, or yield informationthat is easier to work with, than the pointwise values of the function f . This way, functionsappear as continuous linear functionals on some function space. This is how we can generalisethe notion of function. For this chapter, we will assume the point of view of tempered distri-butions, which has a global vision on the whole space R. It is particularly adapted to extendthe Fourier transform to a larger class of maps, beyond the setting of functions in L

1 or L2

seen in the previous chapter. For this first contact with distributions, we keep things simpleby restricting our study to dimension 1.

In the first section, we will define the concept of tempered distributions and see how itgeneralises that of locally integrable functions (modulo a small restriction at infinity). For thispurpose, theorem 8.1.1 will eloquently show how the knowledge of the family

⇣Z

⌦f(x)'(x)dx

⌘

'2E

for a certain vector space E is equivalent to knowing the function in its usual sense (i.e. itsvalues at all points of R). After this, several examples of tempered distributions which are notfunctions in the classical sense will be given. Finally, we will see how the notion of weak-? con-vergence leads us to an adapted sense of convergence for distributions (see proposition 8.1.6).

In the second section, we define the operations that are possible on distributions by trans-posing. This will lead to a striking new fact: all distributions have derivatives (!) and alltempered distributions have a Fourier transform. What follows this is a series of variousformulae that can be quite surprising.

125

In the third section, we show two applications of this theory. The first will involve aclassical analysis problem which is difficult to solve in the strict sense of functions, but whichrequires only a few lines using distributions. Then, we will establish an integral formula whichsolves the equation

u� u00 = f

with u equal to 0 at �1 and +1.

8.1 Definition of tempered distributions and examples

As we explained in the introduction of this chapter, the concept of distributions is based onthe fact that a function can be entirely described by a collection of its weighted means. Wecan thus rethink the notion of a function in terms of linear forms on a function space: thisfunction space will be called the space of test functions. How should we choose this spaceof test functions? If we refer to the L

p spaces we studied, the “smallest” function space weencountered was D(R), the space of smooth functions with compact support. Indeed, if f isin L

2(R) for example, and

8' 2 D(R) ,Z

Rf(x)'(x)dx = 0,

then the function f is equal to zero in L2(R), which means that it is equal to zero almost

everywhere. This shows that D(R) is a good choice for the space of test functions, and this isthe one that applies in the general theory of distributions.

For reasons that will become apparent shortly, we would like the space of test functions tobe stable under the Fourier transform. As we will see, this is not the case for the space D(R)of smooth functions with compact support. Indeed, let ' be a function in D(R). The function

8<

:

C �! C⇣ 7�!

Z

Re�i⇣x

'(x)dx

is well defined because ' has compact support, and the differentiability theorem for functionsof the complex variable defined by integrals yields that it is holomorphic (because it is dif-ferentiable) on C. As the zeros of a holomorphic function are isolated, this function cannotcoincide with a function with compact support on the real line unless this function is zero,which, by the Fourier inversion theorem, yields that ' is zero.

So what properties does the Fourier transform of a function in D(R) have? Corollary 7.1.1on page 120 implies that

sup⇠2Rd

(1 + |⇠|)N��d

d⇠b'(⇠)

�� < 1.

This leads to defining the following space.

Definition 8.1.1. We denote S(R), or simply S, the set of smooth functions f from R to Rsuch that

8n 2 N , kfkn,Sdef= max

kn

supx2R

(1 + |x|)n|f (k)(x)| < 1.

Observe first of all that the space S(R) contains the space D(R) of smooth functions withcompact support. Note also that the Gaussian functions, defined by

x 7�! e�

(x�x0)2

a2

are elements of S.

126

Tempered distributions will appear as linear functionals that satisfy a form of “continuity”property on this space. The following is the precise definition.

Definition 8.1.2. We call a tempered distribution on R a linear functional defined on the

space S(R) that satisfies

9n 2 N , 9C / 8' 2 S(R) , |hT,'i| Ck'kn,S . (8.1)

We denote S 0(R) the set of such linear forms.

We are now going to give some examples of tempered distributions.

Definition 8.1.3. We denote L1M(R) the set of locally integrable functions for which there

exists an integer N such that (1 + |x|)�Nf(x) is in L

1(R).

Exercise 8.1.1. Prove that the spaces Lp(R) are all subsets of L

1M(R).

Exercise 8.1.2. Prove that the functions

x 7�! ex

and x 7�! ex2

are not elements of L1M(R)

Proposition 8.1.1. Let f be a function in L1M(R). The linear form defined by

◆

8<

:

S(R) �! C' 7�!

Z

Rf(x)'(x)dx

is a distribution on R.

Proposition 8.1.2. Let a be a point in R. The linear forms

⇢S(R) �! C� 7�! �(a)

and

8<

:

S(R) �! C� 7�!

X

n2N�(n)

are tempered distributions.

Proposition 8.1.3. Let � be a function on C. Then the function

8<

:R �! C

' 7�! '(x)� '(�x)

x

is integrable, and the linear form defined by

Dvp

1

x, '

E=

1

2

Z1

�1

'(x)� '(�x)

xdx

called the Cauchy principal value of 1/x, is a tempered distribution.

127

Proof. By the mean value inequality, we have��'(x)� '(�x)

�� 2|x| supx2R

|'0(x)|.

By definition of the semi-norms on S, we have��'(x)� '(�x)

�� 2

1 + |x|k'k1,S .

Thus,

8x 2 R / |x| 1,��'(x)� '(�x)

x

�� 2k'k1,S ,

and 8x 2 R / |x| � 1,��'(x)� '(�x)

x

�� 2

|x|2 k'k1,S .

The function x 7�! '(x)� '(�x)

xis therefore integrable on R, and the linear form vp

1

xis

continuous on S. ⇤Remark. Let ' be a test function that is identically zero on a neighbourhood of 0. Then,

Dvp

1

x, '

E=

Z

R

'(x)

xdx.

This means that we have defined an “extension”, of sorts, of the function x�1.

Proposition 8.1.4. Let a be in R and k be in N. The linear form defined by

h�(k)a ,'i def= (�1)k'(k)(a)

is a tempered distribution.

Proof. It suffices to observe that, by definition of the semi-norms on S, we have

|'(k)(a)| Cak'kk,S ,

which yields the result. ⇤

Proposition 8.1.5. The linear form defined by

DPf

1

x2, '

E=

1

2

Z1

�1

'(x) + '(�x)� 2'(0)

x2dx

called the Hadamard finite part of 1/x2, is a tempered distribution.

Proof. The second-order Taylor inequality ensures that

��'(x) + '(�x)� 2'(0)�� |x|2

2supx2R

|'00(x)|.

Thus, we have'(x) + '(�x)� 2'(0)

x2 min

n1,

1

|x|2ok'k2,S .

This yields that Pf1

x2is a tempered distribution. ⇤

128

Remark. If ' is a test function that is identically zero in a neighbourhood of 0, we haveDPf

1

x2, '

E=

Z

R

'(x)

x2dx.

This means that we have defined an “extension” of the function x�2.

The following theorem shows that the space L1M(R) can be identified as a subspace of the

set of tempered distributions.

Theorem 8.1.1. Let ◆ by the map defined by

◆

8<

:

L1M(R) �! S 0(R)f 7�! ◆(f) : ' 7!

Z

Rf(x)'(x)dx

is a linear injection. Moreover, we have the following property: if N is such that (1+|x|)�Nf(x)

is in L1(R), then

|h◆(f),'i| C��(1 + | · |)�N

f��L1k'kN,S .

Proof. Let f be a function in L1M(R) such that ◆(f) = 0. We are going to prove that, for any

pair of real numbers (a, b) with a strictly less than b, we haveZ

b

a

|f(x)|dx = 0

which is enough to prove the theorem. We consider a even function � in D(] � 1, 1[) withintegral equal to 1, and the associated sequence of mollifiers, that is the family

�"(x)def= "

�d�

⇣x

"

⌘·

Finally, we define a function g by

g(x) =f(x)

|f(x)| if f(x) 6= 0, and 0 otherwise.

This function is an element of L1(R). The function

'"def= �" ? (1[a,b]g)

is a smooth function with compact support on R, and therefore a function in S(R). Byassumption, we have

I"def=

Z

Rf(x)'"(x)dx = 0.

For any " in the interval ]0, �[, the function

(x, y) 7�! "�d�

⇣x� y

"

⌘1[a,b](y)g(y)f(x) = "

�d�

⇣x� y

"

⌘1[a,b](y)g(y)f(x)1[a��,b+�](x)

is in L1(R ⇥ R), because it is bounded with compact support. Indeed, the modulus of this

function is bounded by 1, and, by Fubini’s theorem, theorem 5.1.5 on page 83, for every "

smaller than �,Z

R⇥R"�d

��⇣x� y

"

⌘��1[a,b](y)|g(y)|1B(0,�)(x)|f(x)|dxdy k�kL1k1[a��,b+�]fkL1 .

129

The Fubini-Tonelli theorem 5.1.6 on page 84, combined with the evenness of the function �,yields that

I" =

Z

R⇥R�"(x� y)1[a,b](y)g(y)f(x)dxdy

=

Z

R⇥R�"(x� y)1[a,b](y)g(y)1K+B(0,�)(x)f(x)dxdy

=

Z

R

✓Z

R�"(y � x)1[a��,b+�](x)f(x)dx

◆1K(y)g(y)dy

=

Z

R

��" ? (1[a��,b+�])f)

�(y)1[a,b](y)g(y)dy.

Theorem 5.4.5 on page 100 states that

lim"!0

��(�" ? (1[a��,b+�])f)� 1[a��,b+�])f��L1 = 0

As g belongs to L1(R), we have

0 = lim"!0

I"

=

Z1[a��,b+�])(y)f(y)1[a,b](y)g(y)dy

=

Zb

a

f(y)f(y)

|f(y)|dy

=

Zb

a

|f(y)|dy.

The theorem is proved. ⇤Remark. This way, any function in L

1M

can be identified as a tempered distribution, via thefollowing definition.

Definition 8.1.4. We say that a tempered distribution T is a function if and only if there

exists a function f in L1M

such that T = ◆(f).

We shall now define the notion of convergene for a sequence of tempered distributions. Itis a weak-?-type convergence in the sense that we defined in chapter 3.

Definition 8.1.5. Let (Tn)n2N be a sequence of elements of S 0(R), and T be an element

of S 0(R). We say that the sequence (Tn)n2N converges to T if and only if

8' 2 S(R) , limn!1

hTn,'i = hT,'i.

We now give an example of a convergent sequence in this sense, that will show how faraway this notion of convergence for tempered distributions is from the one we use for functions.

Proposition 8.1.6. The sequence of functions SN (x)def=

NX

n=1

sin(nx) is convergent in S 0(R).

Proof. We are looking for a tempered distribution S such that, for any function f in S, wehave

limN!1

NX

n=1

Z

Rsin(nx)'(x)dx = hS,'i.

130

Integrating by parts twice, we get that, for any integer n greater than or equal to 1,Z

Rsin(nx)'(x)dx = � 1

n2

Z

Rsin(nx)'00(x)dx. (8.2)

We define a function ⌃ by

⌃(x) = �1X

n=1

1

n2sin(nx)

This is a continuous, bounded function on R. We then set, for f in S,

hS,'i def=

Z

R⌃(x)'00(x)dx.

This defines a tempered distribution. Moreover, for any function ', by (8.2), we have

limN!1

NX

n=1

Z

Rsin(nx)'(x)dx =

Z

R⌃(x)'00(x)dx = hS,'i.

The proposition is proved. ⇤

8.2 Operations on tempered distributions

The big idea is the following: we have seen that the space S 0 is much “bigger” than the space S;on the functions in S, we can perform a large number of operations, including derivation, theFourier transform, convolution with a function with controlled growth. For the last one, weneed to define the notion of continuous linear map from S to S.

Definition 8.2.1. A linear map A from S to itself is said to be continuous if and only if

8k 2 N , 9(Ck, nk) 2]0,1[⇥N / 8� 2 S , kA�kk,S Ckkfknk,S .

Let is introduce the following two function spaces.

Definition 8.2.2. The space of functions of moderate (or slow) growth, denoted OM , is the

set of functions f that are smooth on R and that satisfy

8k 2 N , 9N 2 N , 9C 2 R+/ 8x 2 R , |f (k)(x)| (1 + |x|)N .

We denote L1S

the space of locally integrable functions that satisfy, for any integer N , that

the function (1 + |x|)Nf(x) is in L1.

Polynomials are excellent examples of functions in OM . Locally integrable functions thatconverge to zero faster than any negative power of |x| as x tends to infinity are excellentexamples of functions in L

1S. For example, the function

x 7�! 1

x↵e�|x|

�,

with ↵ in ]0, 1[ and � positive, belongs to L1S.

Proposition 8.2.1. The following linear maps are continuous from S to S in the sense of

definition 8.2.1:

131

• the map ' 7! '(x)def= '(�x);

• the map ' 7! (�1)k'(k);

• the Fourier transform, as defined in definition 7.1.1 on page 117;

• the map Mf , which denotes multiplication by a function f in OM ;

• the map f?, which denotes convolution by a function f in L1S, i.e.

(f ? ')(x)def=

Z

Rf(y)'(x� y)dy .

Proof. That the first map is continuous is obvious. For the second map, we observe immediatelythat, by definition of the norms k · kk,S ,

k'(`)kk,S Ck'kk+`,S .

To prove that the multiplication map is continuous, it suffices to apply Leibniz’s law, whichstates that

(f')(k) =X

`k

C`

kf(k�`)

'(`).

As the function f belongs to OM , there exists an integer N such that, for any integer ` lessthan k, we have

8x 2 R , |f (`)(x)| C(1 + |x|)N .

We then deduce thatkf'kk,S Cfk'kk+N,S .

Now, we prove the continuity of the Fourier transform. Note that theorem 7.1.1 on page 119yields that

M(b')0 = �iMF(M') = �F�(M')0

�= �b'� F(M'

0),

where M is the multiplication operator by the identity function. As a result,

(1 + |⇠|)|b'(⇠)| 2k'kL1(R) + kM'0kL1(R) 2kfk3,S .

We admit the general proof for the semi-norms with index k on S(R).Finally, we study the convolution operator. Derivating under the integral sign, we have

(f ? ')(k)(x) =

Z

Rf(y)'(k)(x� y)dy.

As |x| 2max{|x� y|, |y|}, we have

(1 + |x|)k 2k⇣(1 + |x� y|)k + (1 + |y|)k

⌘.

Thus, we write

(1 + |x|)k|(f ? ')(k)(x)| 2kZ

R|f(y)|(1 + |x� y|)k|@↵'(x� y)|dy

+ 2kZ

R(1 + |y|)k|f(y)| |'(k)(x� y)|dy.

We then deduce thatkf ? 'kk,S Ck(1 + | · |)kfkL1k'kk,S ,

hence the continuity of convolution. ⇤

132

We can define the transposes of these operations (which often have some of the sameproperties) on the space S 0. These provide spectacular extensions of the notions they represent.All of this relies on the following theorem.

Theorem 8.2.1. Let A be a continuous linear map from S(R) to S(R). Then, the linear

maptA,

tA

⇢S 0(R) �! S 0(R)T 7�! t

AT defined by htAT,'i = hT,A'i.

is well defined. Moreover, it is continuous in the following sense: if (Tn)n2N is a sequence of tem-

pered distributions that converges to a tempered distribution T , then the sequence (tATn)n2Nconverges to

tAT .

Proof. By definition, T is a continuous linear functional on S(R), and A is a continuous linearmap from S(R) to itself. So the composition t

ATdef= T �A is a continuous linear form on S(R),

and is therefore a tempered distribution.Moreover, let (Tn)n2N be a sequence of tempered distributions that converges to T in the

sense of definition 8.1.2, which means that, for any function ✓ in S(R),

limn1

hTn, ✓i = hT, ✓i.

Applying this to ✓ = A', we get that

8' 2 S(R) , limn1

htATn,'i = htAT,'i.

The theorem is proved. ⇤For a first application of this theorem, we extend the operationˇto tempered distributions

by the formula8' 2 S , hT ,'i def

= hT, 'i

Now let us define the derivatives of a tempered distribution.

Definition 8.2.3. Let k be an element of N. The k-th order derivation on S 0is the transpose

of the continuous linear map from S(R) to itself defined by (�1)kdk

dxk, which, in terms of a

tempered distribution T , is written as

8' 2 S(R) , hT (k),'i def

= hT, (�1)k'(k)i.

First note that, for any function f of class C1 with compact support in R, we have, by

integration by parts,

8' 2 S(R) ,Z

Rf0(x)'(x)dx = �

Z

Rf(x)'0(x)dx .

So these two notions of derivation coincide for a function of class C1.

Now, note that the function ⌃ which appeared in the proof of proposition 8.1.6 is noneother than

�⇣d

dx

⌘2 1X

n=1

1

n2sin(n·),

133

which is the second-order derivative in the sense of distributions of the uniformly continuous,bounded function

x 7�! �1X

n=1

1

n2sin(nx).

We are now going to study a very simple example that illustrates how deriving in thedistribution sense detects more kinds of variation than derivation in the classical sense. Indeed,we now know that we can differentiate functions that are not of class C1, for example functionsthat are differentiable everywhere except at one point of R. The following property holds forthe Heaviside function.

Proposition 8.2.2. We have:d

dx1R+ = �0.

Proof. We study, for any given ' in S(R), the integral

�Z

R1R+(x)'(x)dx.


�Z

R1R+(x)'0(x)dx = � lim

A!1

ZA

0'0(x)dx

= limA!1

('(0)� '(A))

= '(0).

Looking back at the definition of the Dirac mass, we see that the proposition is proved. ⇤To further illustrate this notion, we compute the derivative of the function x 7! log |x|.

Proposition 8.2.3. Let vp1

xbe the tempered distribution defined in proposition 8.1.3. In

the sense of distributions, we have the following:

d

dxlog |x| = vp

1

x·

Proof. We study, for any given ' in S(R), the integral

I(') = �Z

Rlog |x|'0(x)dx.

By the dominated convergence theorem, we have

I(') = lim"!0

I"(')def= �

Z

[� 1" ,

1" ]\[�","]

log |x|'0(x)dx.


I"(') = � log "'(�") + log⇣1"

⌘'

⇣�1

"

⌘+

Z�"

�1"

'(x)

xdx

+ log "'(")� log⇣1"

⌘'

⇣1"

⌘+

Z 1"

"

'(x)

xdx

= log "�'(")� '(�")

�� log "'

⇣�1

"

⌘+ log "'

⇣1"

⌘+

Z�"

�1"

'(x)

xdx+

Z 1"

"

'(x)

xdx.

134

By changing the variable, x = �y, we get thatZ

[� 1" ,

1" ]\[�","]

'(x)

xdx = �

Z

[� 1" ,

1" ]\[�","]

'(x)

xdx,

hence

I"(') = log "�'(")� '(�")

�� log "'

⇣�1

"

⌘+ log "'

⇣1"

⌘+

1

2

Z

[� 1" ,

1" ]\[�","]

'(x)� '(�x)

xdx.

As the function ' belongs to S, we have

lim"!0

log "�'(")� '(�")

�� log "'

⇣�1

"

⌘+ log "'

⇣1"

⌘= 0.

Moreover, proposition 8.1.3 states, in particular, that the function

x 7�! '(x)� '(�x)

x

is integrable on R. The dominated convergence theorem then ensures that

�Z

Rlog |x|'0(x)dx =

1

2

Z

R

'(x)� '(�x)

xdx,

which is the statement we set out to prove. ⇤Exercise 8.2.1. Prove that

d

dxvp

1

x= �Pf

1

x2·

We are now going to define the notion of antiderivative (on R) of a tempered distribution.We start by proving the following result.

Theorem 8.2.2. Let T be a tempered distribution such that T0 = 0. Then T is a constant

function, that is

hT,'i = c

Z

R'(x)dx.

Proof. The result relies on the following lemma.

Lemma 8.2.1. Let S0(R) be the set of functions of S(R) with integral equal to 0. The map

P

8<

:

S0(R) �! S(R)

' 7�! x 7!Z

x

�1

'(y)dy

is a continuous linear map from S0(R) to S(R) that satisfies (P�)0 = � for any � in S0(R).

Proof. Let x be less than or equal to �1. We have

|(P')(x)| k'kN+2,S

Zx

�1

1

(1 + |y|)N+2dy

|x|�Nk�kN+2,S

Z 0

�1

1

(1 + |y|)2dy.

If x is greater than or equal to 1, we use the fact that the integral of � is equal to zero towrite that

(P�)(x) = �Z +1

x

�(y)dy,

and we use the same reasoning. We conclude by observing that (P�)0 = �. ⇤

135

Back to the proof of Theorem 8.2.2. This implies that the kernel of T contains the kernel ofthe linear form Z

R�(x)dx

which is T1. Thus T and T1 are proportionnal which means exactly that

hT,�i = c

Z

R�(x)dx

which is exactly the conclusion of Theorem 8.2.2. ⇤

Proposition 8.2.4. Let T be a tempered distribution on R. There exists a tempered distri-

bution S such that S0 = T , and the difference between two distributions satisfying this relation

is constant.

Proof. Let us consider the operator P of Lemma 8.2.1. We kown that, for any � in S wehave P(�0) = �. Let us extend the operator P, which is defined only on the space of functionof S of integral 0 to the whole space S. Let us consider a function �1 in S with integral equalto 1. Let p1 be the projection of S(R) onto S0(R) parallel to the vector line generated by �1,that is the map

p(�) = ��⇣Z +1

�1

�(y)dy⌘�1. (8.3)

Because (�0)) = �, by Lemma 8.2.1, we have P(�0) = � which can be written

P � d

dx= IdS(R) .

By transposing, we deduce that

t

⇣P � d

dx

⌘= IdS0(R) .

As t(AB) = tB

tA, we get that

t

⇣d

dx

⌘� tP = IdS0(R) .

The fact that t

⇣d

dx

⌘= � d

dxyields the result. ⇤

As we assume T0 to be zero, we deduce that

hT, p(')i = 0 = hT,'i �⇣Z +1

�1

'(y)dy⌘hT,'1i.

The theorem is proved.

136

We now apply these ideas in order to find antiderivatives of distributions in dimension 1.Let '1 be a function in S(R) with integral equal to 1. We will show that S

def= �tPT

satisfies S0 = T . We observe that, for any function ' in S(R), the integral of '0 is equal to 0,

thusP � d

dx= IdS(R) .

By transposing, we deduce that

t

⇣P � d

dx

⌘= IdS0(R) .

As t(AB) = tB

tA, we get that

t

⇣d

dx

⌘� tP = IdS0(R) .

The fact that t

⇣d

dx

⌘= � d

dxyields the result. ⇤

Remark. The distribution S can be written

hS,�i def= �hT,P(�)i with (P(�))(x)

def=

Zx

�1

⇣�(y)�

⇣Z 1

�1

�(t)dt⌘�1(y)

⌘dy,

where �1 is a function in S(R) with integral equal to 1, for examplep⇡e

�x2 .

The relation between derivative and antiderivative can be precisely described in the fol-lowing theorem.

Theorem 8.2.3. Let T be a tempered distribution on R. If its derivative (in the sense of

distributions) T0is a function in L

1M(R), then T is a continuous function, and we have

T (x) =

Zx

0T0(y)dy + C.

Proof. The proof relies on the following lemma that we admit for the moment.

Lemma 8.2.2. Let f be a function in L1M(R). Then, in the sense of distributions, we have

d

dx

Zx

0f(y)dy = f(x)

Set F (x) =

Zx

a

T0(y)dy. By Lemma 8.2.2, we have F

0 = T0. By Proposition 8.2.2, we also

haveT � F = C.

This yields the theorem, providing we can prove lemma 8.2.2 of course. ⇤Proof of Lemma 8.2.2. By definition of derivation in the distribution sense, we need to provethat if

F (x)def=

Zx

0f(y)dy,

then we have8' 2 S(R) , �

Z

RF (x)'0(x)dx =

Z

Rf(x)'(x)dx.

137

Observe first that if

I+ def=�(x, y) 2 R2

, 0 y x

and I� def=�(x, y) 2 R2

, x y 0 ,

then the functions1I+(x, y)f(y)'0(x) and 1I�(x, y)f(y)'0(x)

are integrable on R2. Indeed, by definition of L1M(R),

I(f,')def=

Z

R21I+(x, y)|f(y)| |'0(x)|dxdy

Z

R21I+(x, y)(1 + |y|)M | ef(y)| |'0(x)|dxdy with ef 2 L

1(R).

By Fubini’s theorem for non-negative functions, we have

I(f,') CM,a

Z

R(1 + |x|)M+1|'0(x)|dx CM,ak'kM+1,S .

The same bound for 1I�(x, y)f(y)'0(x) is proved the same way. Note that, thanks to Fubini’stheorem, we can write that

�Z

RF (x)'0(x)dx = �

Z

R

⇣Z 1

y

'0(x)dx

⌘1[0,1[(y)f(y)dy

+

Z

R

⇣Z y

1

'0(x)dx

⌘1]�1,0](y)f(y)dy

=

Z

R'(y)1[0,1[(y)f(y)dy +

Z

R'(y)1]�1,0](y)f(y)dy

=

Z

R'(y)f(y)dy.

This ends the proof of lemma 8.2.2, and therefore also that of theorem 8.2.3. ⇤Now we see how to multiply a tempered distribution by a function in OM .

Definition 8.2.4. Let ✓ be a function in OM . The ✓ multiplier operator on S 0is the transpose

of the ✓ multiplier operator on S, which means that, for T in S 0,

h✓T,'i def= hT, ✓'i.

It is obvious that if T is a function in L1M

, then multiplication in this sense coincides withthe usual product in the function sense.

We prove a generalisation of Leibniz’s law.

Proposition 8.2.5. If ✓ belongs to OM and T belongs to S 0, then we have, for any integer k,

(✓T )(k) =X

`k

C`

k✓(k�`)

T(`).

Proof. As in the classical function setting, we only need to prove that (✓T )0 = ✓0T + ✓T

0,formula that we then iterate. By definition of derivation and multiplication by ✓, we have

h(✓T )0,'i = �h✓T,'0i= �hT, ✓'0i.

138

Leibniz’s law for smooth functions implies that

h(✓T )0,'i = �hT, (✓')0i+ hT, ✓0'i.

Again, by definition of derivation and multiplication, we deduce that

h(✓T )0,'i = hT 0, ✓'i+ hT, ✓0'i

= h✓T 0 + ✓0T,'i.

The proposition is proved. ⇤Now we define the convolution of a tempered distribution with a function in L

1S.

Definition 8.2.5. Let f be a function in L1S. We define the convolution of f and tempered

distribution as the transpose of the convolution of f and an element of S, which means

hf ? T,'i def= hT, f ? 'i.

Let us check that we extend the notion of convolution defined in theorem 5.4.1 on page 97.Let g be a function in L

1(R). We have

h◆(g), f ? 'i =

Z

Rg(x)(f ? ')(x)dx

=

Z

Rg(x)

⇣Z

Rf(x� y)'(y)dy

⌘dx.

Fubini’s theorem implies that

h◆(g), f ? 'i =

Z

R⇥Rg(x)f(y � x)'(x)dx

=

Z

Rg(x)

⇣Z

Rf(x� y)'(y)dy

⌘dx

=

Z

R(g ? f)(y)'(y)dy,

hence the result.

The relation between convolution and derivation is described by the following proposition.

Proposition 8.2.6. Let f be in L1S, and T be in S 0

. For any integer k, we have

(f ? T )(k) = f ? T(k)

.

Proof. By definition of derivation and convolution on S 0, we have

h(f ? T )(k),'i = hf ? T, (�1)k'(k)i= hT, f ? (�1)k'(k)i.

By the differentiation theorem for functions defined by integrals, we have

f ? '(k) = (f ? ')(k).

Once again using the definition of derivation and convolution on S 0, we deduce that

h(f ? T )(k),'i = hT (k), f ? 'i

= hf ? T (k),'i,

hence the proposition. ⇤

139

A fundamental and valuable point in this theory is the extension of the Fourier transformto tempered distributions. In the previous chapter, we saw that the Fourier transform wasoriginally defined on L

1(R), but could be extended to L2(R) by density. The fact that the

space S is invariant under Fourier transform (theorem 7.2.1 on page 121) allows us to extendthe Fourier transform to S 0(R) by duality.

Definition 8.2.6. The Fourier transform on S 0is the transpose of the Fourier transform on S,

that is

8' 2 S , hFS0T,'i = hT, b'i.

The following proposition shows that this is indeed an extension of the Fourier transform.

Proposition 8.2.7. For any function f in L1(R), we have

FS0

�◆(f)

�= ◆( bf).

Proof. Reprising the notations used in theorem 8.1.1, we have, for any function ' in S, that

hFS0◆(f),'i = h◆(f), b'i

=

Z

Rf(x)b'(x)dx.

Given that the function(x, y) 7�! f(x)e�i(x|⇠)

'(⇠)

is integrable on R⇥ R, we deduce that

hFS0◆(f),'i =

Z

Rbf(⇠)'(⇠)d⇠

= h◆( bf),'i,

hence the proposition. ⇤We now generalise the basic formulae on the Fourier transform to tempered distributions.

Theorem 8.2.4. For any tempered distribution T , we have

F2S0T = 2⇡T ,

�FS0(T )

�0= �iFS0(MT ) and FS0(T 0) = iMFS0T.

Proof. For any ' in S, we have F2' = 2⇡'. Transposing this relation, we have, for any T

in S 0,(tF)2T = F2

S0T = 2⇡T ,

which yields the first formula. For ' in S, we know by theorem 7.1.1 that

F(i'0) = M b' and F(M') = �id

d⇠b'.

The last two equalities are obtained by transposing these. ⇤Remark. Henceforth, we no longer use different notations for the Fourier transform on func-tions in L

1(R) and the one on tempered distributions. So, for any T in S 0(R), we indifferentlydenote its Fourier transform bT or F(T ).

140

Some applications of the above theorem 8.2.4 consist of computations of Fourier transformsof functions that are not in L

1. For example, let us consider the constant function 1, and thetempered distribution vp

1

xfrom proposition 8.1.3.

Proposition 8.2.8. We have the following formulae:

F(�0) = 1 , F(1) = 2⇡�0 and F⇣vp

1

x

⌘(⇠) = �i⇡

⇠

|⇠| ·

Proof. The first formula is a simple consequence of the fact that

hF(�0),'i = b'(0) =Z

R'(x)dx.

The second one comes from the Fourier inversion formula on S 0, which is

F2�0 = F(1) = 2⇡�0.

Let us now take a close look at the principal value of 1/x. By definition of this distribution,for any function ' in S(R), we have

Dvp

1

x, x'

E=

1

2

Z1

�1

x'(x) + x'(�x)

xdx

=1

2

Z

R

�'(x) + '(�x)

�dx

=

Z

R'(x)dx

= h1,'i.

Thus, x vp1

x= 1. Using theorem 8.2.4, we deduce that

�1

i

d

d⇠F⇣vp

1

x

⌘= F(1).

But we know that F1 = 2⇡�0, so we have

d

d⇠

bT = �2i⇡�0.

Thus, by proposition 8.2.2 and lemma 8.2.2, we have

bT = �2i⇡H(⇠) + C,

where C is a scalar constant. To get the value of this constant, we observe that the principalvalue of 1/x is an odd distribution, in the sense that

ˇvp

1

x= �vp

1

x·

We let the reader prove that the Fourier transform of an odd tempered distribution is an oddtempered distribution. Thus, the distribution bT must be odd, so the constant must be C = i⇡.The result is proved. ⇤

141

We now study the relations between convolution, multiplication and the Fourier trans-form, with an aim to extend (i.e. transpose) the result of proposition 7.1.4 on page 121 andcorollary 7.2.2 on page 123.

Theorem 8.2.5. Let ✓ be a function in S, and T be a tempered distribution. We have

F(✓ ? T ) = b✓ bT and F(✓T ) = (2⇡)�1b✓ ? bT .

Proof. By definition of the Fourier transform and of convolution, we have, for any function 'in S,

hF(✓ ? T ),'i = h✓ ? T, b'i= hT, ✓ ? b'i= h bT ,F�1(✓ ? b')i.

By the Fourier inversion theorem and the formula for computing Fourier transforms of theconvolution of two functions (see proposition 7.1.4 on page 121), we have

F�1(✓ ? b')(⇠) = (2⇡)�1F(✓ ? b')(�⇠)

= (2⇡)�1Z

R⇥Rei(x|⇠)

✓(y � x)b'(y)dydx

= (2⇡)�1Z

R⇥Re�i(y�x|⇠)

✓(y � x)ei(y|⇠) b'(y)dydx

= b✓(⇠)F�1(b')(⇠)= b✓(⇠)'(⇠).

By definition of multiplication for tempered distributions, we get that

hF(T ? ✓),'i = h bT , b✓'i = hb✓ bT ,'i,

which yields the first formula. To establish the second one, we apply the first to F(✓) and F(T ),so we get

F�F(✓) ? F(T )

�= F2

✓F2T = (2⇡)2✓T.

Taking the Fourier transform of this equality yields

F2(F(✓) ? F(T )�= (2⇡)2F(✓T ),

and the Fourier inversion formula ensures the result. ⇤

8.3 Two applications

The following theorem is an application of the previous result. It solves a classical harmonicanalysis problem.

Theorem 8.3.1. There exists a constant C such that, for any function ' in S(R), we have

��' ? vp1

x

��L2

= ⇡k'kL2 .

142

Proof. We use the Fourier transform. By proposition 8.2.5, we have

F⇣' ? vp

1

x

⌘= b'F

⇣vp

1

x

⌘= �i⇡

⇠

|⇠| b'(⇠).

The Fourier-Plancherel relation in theorem 7.2.1 on page 121 ensures that��' ? vp

1

x

��L2

= (2⇡)12

��F⇣' ? vp

1

x

⌘��L2

= (2⇡)12

��i⇡⇠

|⇠| b'(⇠)��L2

= ⇡k'kL2 ,

which proves the theorem. ⇤Let us show an application of this theory to get an explicit solution to a differential equa-

tion.

Theorem 8.3.2. Let T be a tempered distribution. There exists a unique tempered distri-

bution S that solves

S � S00 = T

in the space S 0(R). This solution is given by the formula

S =1

2e�|·|

? T .

Proof. To prove this, we use Theorem 8.2.4 to state that

F(S � S00) = (1 + |⇠|2)bS.

The Fourier inversion formula in Theorem 8.2.4 then yields

S � S00 = T () (1 +M

2)bS = bT .

The function ⇠ 7! (1 + |⇠|2)±1 is a smooth function of slow growth, hence

bS =1 + |⇠|21 + |⇠|2

bT

=1

1 + |⇠|2bT .

So the solution of the equation can be written as

S = F�1

✓1

1 + |⇠|2bT◆.

Proposition 7.1.1 on page 117, which was

1

2F(e�|·|)(⇠) =

1

1 + |⇠|2 ,

ends the proof of the theorem. ⇤

143

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